The Power Integrations of Trigonometric and Hyperbolic Functions
Abstract: For importance of the trigonometric integrals, we have in this paper finding a series of power, some of trigonometric functions that did not exist before in the first section. As shown in the Section two, where, the integration of trigono-metric function with power n has been achieved and approved, this result is considered as the first achievement. while in the third section we find integrals for multiplying of trigonometric functions with powers n and m. Finally, in Section 4, we find series of power of hyperbolic functions, integrals of hyper-bolic functions with powers n and integrals for multiplying of hyperbolic func-tions with powers n and m. 1. Introduction

We follow the terminology of    . The Trigonometry had resulted from the continuous interaction between mathematics and astronomy, and this remained the special relationship between trigonometry and astronomy until the third century AD, when it began to disconnect Nasir al-Din al-Tusi (1201-1274) AD. In the middle of the seventeenth century when calculus has been developed by Issac Newton and that is done by inventing new form and relationship between mathematics and physical phenomena, moreover. Newton’s work proved many functions as a series of infinity. with respect x and then Newton obtained series since x and similar series of cosine of x as well as tangent x with the invention of calculus, and re-considered the trigonometric functions where still play an important role in both pure and applied mathematics analysis   .

Theorem 1.1  : For all $k\in N$. then:

1) ${\mathrm{sin}}^{2k-1}\left(x\right)=\frac{{\left(-1\right)}^{k-1}}{{2}^{2k-2}}\underset{n=1}{\overset{k}{\sum }}{\left(-1\right)}^{n-1}\left(\begin{array}{c}2k-1\\ n-1\end{array}\right)\mathrm{sin}\left(2k-2n+1\right)x$.

2) ${\mathrm{sin}}^{2k}\left(x\right)=\frac{1}{{2}^{2k}}\left(\begin{array}{c}2k\\ k\end{array}\right)+\frac{{\left(-1\right)}^{k}}{{2}^{2k-1}}\underset{n=0}{\overset{k-1}{\sum }}{\left(-1\right)}^{n}\left(\begin{array}{c}2k\\ n\end{array}\right)\mathrm{cos}\left(2k-2n\right)x$.

3) ${\mathrm{cos}}^{2k-1}\left(x\right)=\frac{1}{{2}^{2k-2}}\underset{n=1}{\overset{k}{\sum }}\left(\begin{array}{c}2k-1\\ n-1\end{array}\right)\mathrm{cos}\left(2k-2n+1\right)x$.

4) ${\mathrm{cos}}^{2k}\left(x\right)=\frac{1}{{2}^{2k}}\left(\begin{array}{c}2k\\ k\end{array}\right)+\frac{1}{{2}^{2k-1}}\underset{n=0}{\overset{k-1}{\sum }}\left(\begin{array}{c}2k\\ n\end{array}\right)\mathrm{cos}\left(2k-2n\right)x$.#

Theorem 1.2: For all $k\in N$. then:

${\mathrm{tan}}^{2k}\left(x\right)={\left(-1\right)}^{k}+{\mathrm{sec}}^{2}\left(x\right)\underset{n=1}{\overset{k}{\sum }}{\left(-1\right)}^{n+1}{\mathrm{tan}}^{2k-2n}\left(x\right)$.span class="bracketMark">(1.2.1)

${\mathrm{tan}}^{2k+1}\left(x\right)={\left(-1\right)}^{k}\mathrm{tan}\left(x\right)+{\mathrm{sec}}^{2}\left(x\right)\underset{n=1}{\overset{k}{\sum }}{\left(-1\right)}^{n+1}{\mathrm{tan}}^{2k-2n+1}\left(x\right)$.span class="bracketMark">(1.2.2)

${\mathrm{cot}}^{2k}\left(x\right)={\left(-1\right)}^{k}+{\mathrm{csc}}^{2}\left(x\right)\underset{n=1}{\overset{k}{\sum }}{\left(-1\right)}^{n+1}{\mathrm{cot}}^{2k-2n}\left(x\right)$.span class="bracketMark">(1.2.3)

${\mathrm{cot}}^{2k+1}\left(x\right)={\left(-1\right)}^{k}\mathrm{cot}\left(x\right)+{\mathrm{csc}}^{2}\left(x\right)\underset{n=1}{\overset{k}{\sum }}{\left(-1\right)}^{n+1}{\mathrm{cot}}^{2k-2n+1}\left(x\right)$.span class="bracketMark">(1.2.4)

Proof:

1) We use mathematical induction on k, $k\ge 1$.

For $k=1$. we find by direct:

${\mathrm{tan}}^{2}\left(x\right)=-1+{\mathrm{sec}}^{2}\left(x\right)$.

For $k=2$. then

$\begin{array}{c}{\mathrm{tan}}^{4}\left(x\right)={\mathrm{tan}}^{2}\left(x\right)\left(-1+{\mathrm{sec}}^{2}\left(x\right)\right)\\ =1-{\mathrm{sec}}^{2}\left(x\right)+{\mathrm{tan}}^{2}\left(x\right){\mathrm{sec}}^{2}\left(x\right)\\ =1+{\mathrm{sec}}^{2}\left(x\right)\left({\mathrm{tan}}^{2}\left(x\right)-1\right)\\ ={\left(-1\right)}^{2}+{\mathrm{sec}}^{2}\left(x\right)\underset{n=1}{\overset{2}{\sum }}{\left(-1\right)}^{n+1}{\mathrm{tan}}^{4-2n}\left(x\right)\end{array}$

We assume that (1.2.1) is true for $k=m$. $m\ge 1$.nd prove that is true for $k=m+1$.

${\mathrm{tan}}^{2\left(m+1\right)}\left(x\right)={\mathrm{tan}}^{2m}\left(x\right)\cdot {\mathrm{tan}}^{2}\left(x\right)=-{\mathrm{tan}}^{2m}\left(x\right)+{\mathrm{tan}}^{2m}\left(x\right){\mathrm{sec}}^{2}\left(x\right)$

For hypotheses induction, we get:

$\begin{array}{l}{\mathrm{tan}}^{2\left(m+1\right)}\left(x\right)\\ =-\left[{\left(-1\right)}^{m}+{\mathrm{sec}}^{2}\left(x\right)\underset{n=1}{\overset{m}{\sum }}{\left(-1\right)}^{n+1}{\mathrm{tan}}^{2m-2n}\left(x\right)\right]+{\mathrm{sec}}^{2}\left(x\right){\mathrm{tan}}^{2m}\left(x\right)\\ ={\left(-1\right)}^{m+1}+{\mathrm{sec}}^{2}\left(x\right)\underset{n=1}{\overset{m+1}{\sum }}{\left(-1\right)}^{n+1}{\mathrm{tan}}^{2\left(m+1\right)-2n}\left(x\right)\end{array}$

Therefore, the relation is true for all k, $k\ge 1$.

2) Also use mathematical induction on k, $k\ge 1$.

For $k=1$. we find that:

${\mathrm{tan}}^{3}\left(x\right)=\mathrm{tan}\left(x\right)\left(-1+{\mathrm{sec}}^{2}\left(x\right)\right)=-\mathrm{tan}\left(x\right)+\mathrm{tan}\left(x\right){\mathrm{sec}}^{2}\left(x\right)$

For $k=2$. then

$\begin{array}{c}{\mathrm{tan}}^{5}\left(x\right)={\mathrm{tan}}^{3}\left(x\right)\left(-1+{\mathrm{sec}}^{2}\left(x\right)\right)\\ =-{\mathrm{tan}}^{3}\left(x\right)+{\mathrm{tan}}^{3}\left(x\right){\mathrm{sec}}^{2}\left(x\right)\\ =\mathrm{tan}\left(x\right)-{\mathrm{sec}}^{2}\left(x\right)\mathrm{tan}\left(x\right)+{\mathrm{sec}}^{2}\left(x\right){\mathrm{tan}}^{3}\left(x\right)\\ =\mathrm{tan}\left(x\right)+{\mathrm{sec}}^{2}\left(x\right)\left({\mathrm{tan}}^{3}\left(x\right)-\mathrm{tan}\left(x\right)\right)\\ ={\left(-1\right)}^{2}\mathrm{tan}\left(x\right)+{\mathrm{sec}}^{2}\left(x\right)\underset{n=1}{\overset{2}{\sum }}{\left(-1\right)}^{n+1}{\mathrm{tan}}^{5-2n}\left(x\right)\end{array}$

We assume that (1.2.2) is true for $k=m$. $m\ge 1$.nd prove that is true for $k=m+1$.

${\mathrm{tan}}^{2\left(m+1\right)+1}\left(x\right)={\mathrm{tan}}^{2m+1}\left(x\right)\cdot {\mathrm{tan}}^{2}\left(x\right)=-{\mathrm{tan}}^{2m+1}\left(x\right)+{\mathrm{tan}}^{2m+1}\left(x\right){\mathrm{sec}}^{2}\left(x\right)$

For hypotheses induction, we get:

$\begin{array}{l}{\mathrm{tan}}^{2\left(m+1\right)+1}\left(x\right)\\ =-\left[{\left(-1\right)}^{m}\mathrm{tan}\left(x\right)+{\mathrm{sec}}^{2}\left(x\right)\underset{n=1}{\overset{m}{\sum }}{\left(-1\right)}^{n+1}{\mathrm{tan}}^{2m-2n+1}\left(x\right)\right]+{\mathrm{sec}}^{2}\left(x\right){\mathrm{tan}}^{2m+1}\left(x\right)\\ ={\left(-1\right)}^{m+1}\mathrm{tan}\left(x\right)+{\mathrm{sec}}^{2}\left(x\right)\underset{n=1}{\overset{m+1}{\sum }}{\left(-1\right)}^{n+1}{\mathrm{tan}}^{2\left(m+1\right)+1-2n}\left(x\right)\end{array}$

Therefore, the relation is true for all k, $k\ge 1$.

3) We can prove it with the same method.

4) We can prove it with the same method. #

Theorem 1.3: For all $k\in N$. then:

${\mathrm{sec}}^{2k}\left(x\right)={\mathrm{sec}}^{2}\left(x\right)\underset{n=0}{\overset{k-1}{\sum }}\left(\begin{array}{c}k-1\\ n\end{array}\right){\mathrm{tan}}^{2n}\left(x\right)$. (1.3.1)

${\mathrm{csc}}^{2k}\left(x\right)={\mathrm{csc}}^{2}\left(x\right)\underset{n=0}{\overset{k-1}{\sum }}\left(\begin{array}{c}k-1\\ n\end{array}\right){\mathrm{cot}}^{2n}\left(x\right)$. (1.3.2)

Proof:

1) $\begin{array}{c}{\mathrm{sec}}^{2k}\left(x\right)={\mathrm{sec}}^{2}\left(x\right)\cdot {\left({\mathrm{sec}}^{2}\left(x\right)\right)}^{k-1}={\mathrm{sec}}^{2}\left(x\right)\cdot {\left(1+{\mathrm{tan}}^{2}\left(x\right)\right)}^{k-1}\\ ={\mathrm{sec}}^{2}\left(x\right)\underset{n=0}{\overset{k-1}{\sum }}\left(\begin{array}{c}k-1\\ n\end{array}\right){\mathrm{tan}}^{2n}\left(x\right)\end{array}$.by binomial theory).

2) $\begin{array}{c}{\mathrm{csc}}^{2k}\left(x\right)={\mathrm{csc}}^{2}\left(x\right)\cdot {\left({\mathrm{csc}}^{2}\left(x\right)\right)}^{k-1}={\mathrm{csc}}^{2}\left(x\right)\cdot {\left(1+{\mathrm{cot}}^{2}\left(x\right)\right)}^{k-1}\\ ={\mathrm{csc}}^{2}\left(x\right)\underset{n=0}{\overset{k-1}{\sum }}\left(\begin{array}{c}k-1\\ n\end{array}\right){\mathrm{cot}}^{2n}\left(x\right)\end{array}$.by binomial theory).

2. The Integration of Trigonometric Function Power

In this section, we find the integrations of power of function trigonometric

Theorem 2.1: For all $k\in N$. then:

1) $\int {\mathrm{sin}}^{2k-1}\left(x\right)\text{d}x=\frac{{\left(-1\right)}^{k-1}}{{2}^{2k-2}}\underset{n=1}{\overset{k}{\sum }}\frac{{\left(-1\right)}^{n-1}\left(\begin{array}{c}2k-1\\ n-1\end{array}\right)}{2k-2n+1}\mathrm{cos}\left(2k-2n+1\right)x+c$

2) $\int {\mathrm{sin}}^{2k}\left(x\right)\text{d}x=\frac{1}{{2}^{2k}}\left(\begin{array}{c}2k\\ k\end{array}\right)x+\frac{{\left(-1\right)}^{k}}{{2}^{2k-1}}\underset{n=0}{\overset{k-1}{\sum }}\frac{{\left(-1\right)}^{n}\left(\begin{array}{c}2k\\ n\end{array}\right)}{2k-2n}\mathrm{sin}\left(2k-2n\right)x+c$

3) $\int {\mathrm{cos}}^{2k-1}\left(x\right)\text{d}x=\frac{1}{{2}^{2k-2}}\underset{n=1}{\overset{k}{\sum }}\frac{\left(\begin{array}{c}2k-1\\ n-1\end{array}\right)}{2k-2n+1}\mathrm{sin}\left(2k-2n+1\right)x+c$.

4) $\int {\mathrm{cos}}^{2k}\left(x\right)\text{d}x=\frac{1}{{2}^{2k}}\left(\begin{array}{c}2k\\ k\end{array}\right)x+\frac{1}{{2}^{2k-1}}\underset{n=0}{\overset{k-1}{\sum }}\frac{\left(\begin{array}{c}2k\\ n\end{array}\right)}{2k-2n}\mathrm{sin}\left(2k-2n\right)x+c$. #

Proof: Directed from Theorem (1.1). #

Theorem 2.2: For all $k\in N$. then:

1) $\int {\mathrm{tan}}^{2k}\left(x\right)\text{d}x={\left(-1\right)}^{k}x+\underset{n=1}{\overset{k}{\sum }}\frac{{\left(-1\right)}^{n+1}}{2k-2n+1}{\mathrm{tan}}^{2k-2n+1}\left(x\right)+c$.

2) $\int {\mathrm{tan}}^{2k+1}\left(x\right)\text{d}x={\left(-1\right)}^{k+1}\mathrm{ln}|\mathrm{cos}\left(x\right)|+\underset{n=1}{\overset{k}{\sum }}\frac{{\left(-1\right)}^{n+1}}{2k-2n+2}{\mathrm{tan}}^{2k-2n+2}\left(x\right)+c$

3) $\int {\mathrm{cot}}^{2k}\left(x\right)\text{d}x={\left(-1\right)}^{k}x+\underset{n=1}{\overset{k}{\sum }}\frac{{\left(-1\right)}^{n}}{2k-2n+1}{\mathrm{cot}}^{2k-2n+1}\left(x\right)+c$.

4) $\int {\mathrm{cot}}^{2k+1}\left(x\right)\text{d}x={\left(-1\right)}^{k}\mathrm{ln}|\mathrm{sin}\left(x\right)|+\underset{n=1}{\overset{k}{\sum }}\frac{{\left(-1\right)}^{n}}{2k-2n+2}{\mathrm{cot}}^{2k-2n+2}\left(x\right)+c$

5) $\int {\mathrm{sec}}^{2k}\left(x\right)\text{d}x=\underset{n=0}{\overset{k-1}{\sum }}\frac{\left(\begin{array}{c}k-1\\ n\end{array}\right)}{2n+1}{\mathrm{tan}}^{2n+1}\left(x\right)+c$.

6) $\int {\mathrm{csc}}^{2k}\left(x\right)\text{d}x=-\underset{n=0}{\overset{k-1}{\sum }}\frac{\left(\begin{array}{c}k-1\\ n\end{array}\right)}{2n+1}{\mathrm{cot}}^{2n+1}\left(x\right)+c$.

Proof: Directed from Theorem (1.2) and Theorem (1.3). #

Lemma 2.3: For all $k\in N$. then:

$\int {\mathrm{sec}}^{2k+1}\left(x\right)\text{d}x=\frac{1}{2k}\left[{\mathrm{sec}}^{2k-1}\left(x\right)\mathrm{tan}\left(x\right)+\left(2k-1\right)\int {\mathrm{sec}}^{2k-1}\left(x\right)\text{d}x\right]$

Proof: $\int {\mathrm{sec}}^{2k+1}\left(x\right)\text{d}x=\int {\mathrm{sec}}^{2k-1}\left(x\right){\mathrm{sec}}^{2}\left(x\right)\text{d}x$

Using integration by parts:

Let $u={\mathrm{sec}}^{2k-1}\left(x\right)$. then $\text{d}u=\left(2k-1\right){\mathrm{sec}}^{2k-1}\left(x\right)\mathrm{tan}\left(x\right)\text{d}x$

$\text{d}v={\mathrm{sec}}^{2}\left(x\right)\text{d}x$. then $v=\mathrm{tan}\left(x\right)$

$\begin{array}{l}\therefore \int {\mathrm{sec}}^{2k+1}\left(x\right)\text{d}x\\ ={\mathrm{sec}}^{2k-1}\left(x\right)\mathrm{tan}\left(x\right)-\left(2k-1\right)\int {\mathrm{sec}}^{2k-1}\left(x\right){\mathrm{tan}}^{2}\left(x\right)\text{d}x\\ ={\mathrm{sec}}^{2k-1}\left(x\right)\mathrm{tan}\left(x\right)-\left(2k-1\right)\int {\mathrm{sec}}^{2k-1}\left(x\right)\left({\mathrm{sec}}^{2}\left(x\right)-1\right)\text{d}x\\ ={\mathrm{sec}}^{2k-1}\left(x\right)\mathrm{tan}\left(x\right)-\left(2k-1\right)\int {\mathrm{sec}}^{2k+1}\left(x\right)\text{d}x+\left(2k-1\right)\int {\mathrm{sec}}^{2k-1}\left(x\right)\text{d}x\end{array}$

$\therefore 2k\int {\mathrm{sec}}^{2k+1}\left(x\right)\text{d}x={\mathrm{sec}}^{2k-1}\left(x\right)\mathrm{tan}\left(x\right)+\left(2k-1\right)\int {\mathrm{sec}}^{2k-1}\left(x\right)\text{d}x$.

Hence

$\int {\mathrm{sec}}^{2k+1}\left(x\right)\text{d}x=\frac{1}{2k}\left[{\mathrm{sec}}^{2k-1}\left(x\right)\mathrm{tan}\left(x\right)+\left(2k-1\right)\int {\mathrm{sec}}^{2k-1}\left(x\right)\text{d}x\right]$.#

Note: From clearly that:

$\int {\mathrm{sec}}^{3}\left(x\right)\text{d}x=\frac{1}{2}\left[\mathrm{sec}\left(x\right)\mathrm{tan}\left(x\right)+\mathrm{ln}|\mathrm{sec}\left(x\right)+\mathrm{tan}\left(x\right)|\right]+c$.

Theorem 2.4: For all $k\in N$. then:

$\begin{array}{c}\int {\mathrm{sec}}^{2k+1}\left(x\right)\text{d}x=\frac{\underset{i=1}{\overset{k-1}{\prod }}\left(2i+1\right)}{\underset{i=1}{\overset{k}{\prod }}\left(2i\right)}\mathrm{ln}|\mathrm{sec}\left(x\right)+\mathrm{tan}\left(x\right)|+\frac{1}{2k}{\mathrm{sec}}^{2k-1}\left(x\right)\mathrm{tan}\left(x\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+\mathrm{tan}\left(x\right)\underset{j=1}{\overset{k-1}{\sum }}\frac{\underset{i=j}{\overset{k-1}{\prod }}\left(2i+1\right)}{\underset{i=j}{\overset{k}{\prod }}\left(2i\right)}{\mathrm{sec}}^{2i-1}\left(x\right)+c\end{array}$

Proof: Since

$\int {\mathrm{sec}}^{2k+1}\left(x\right)\text{d}x=\frac{1}{2k}\left[{\mathrm{sec}}^{2k-1}\left(x\right)\mathrm{tan}\left(x\right)+\left(2k-1\right)\int {\mathrm{sec}}^{2k-1}\left(x\right)\text{d}x\right]$

(by Lemma 2.3). Then

$\int {\mathrm{sec}}^{2k-1}\left(x\right)\text{d}x=\frac{1}{2k-2}\left[{\mathrm{sec}}^{2k-3}\left(x\right)\mathrm{tan}\left(x\right)+\left(2k-3\right)\int {\mathrm{sec}}^{2k-3}\left(x\right)\text{d}x\right]$

and $\int {\mathrm{sec}}^{2k-3}\left(x\right)\text{d}x=\frac{1}{2k-4}\left[{\mathrm{sec}}^{2k-5}\left(x\right)\mathrm{tan}\left(x\right)+\left(2k-5\right)\int {\mathrm{sec}}^{2k-5}\left(x\right)\text{d}x\right]$.

and so on

$\int {\mathrm{sec}}^{5}\left(x\right)\text{d}x=\frac{1}{4}\left[{\mathrm{sec}}^{3}\left(x\right)\mathrm{tan}\left(x\right)+3\int {\mathrm{sec}}^{3}\left(x\right)\text{d}x\right]$

and $\int {\mathrm{sec}}^{3}\left(x\right)\text{d}x=\frac{1}{2}\left[\mathrm{sec}\left(x\right)\mathrm{tan}\left(x\right)+\mathrm{ln}|\mathrm{sec}\left(x\right)+\mathrm{tan}\left(x\right)|\right]+c$.

$\begin{array}{l}\therefore \int {\mathrm{sec}}^{2k+1}\left(x\right)\text{d}x\\ =\frac{1}{2k}{\mathrm{sec}}^{2k-1}\left(x\right)\mathrm{tan}\left(x\right)+\frac{2k-1}{2k\left(2k-2\right)}{\mathrm{sec}}^{2k-3}\left(x\right)\mathrm{tan}\left(x\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }+\frac{\left(2k-1\right)\left(2k-3\right)}{2k\left(2k-2\right)\left(2k-4\right)}{\mathrm{sec}}^{2k-5}\left(x\right)\mathrm{tan}\left(x\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }+\frac{\left(2k-1\right)\left(2k-3\right)\left(2k-5\right)}{2k\left(2k-2\right)\left(2k-4\right)\left(2k-6\right)}{\mathrm{sec}}^{2k-7}\left(x\right)\mathrm{tan}\left(x\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }+\frac{\left(2k-1\right)\left(2k-3\right)\left(2k-5\right)\left(2k-7\right)}{2k\left(2k-2\right)\left(2k-4\right)\left(2k-6\right)\left(2k-8\right)}{\mathrm{sec}}^{2k-9}\left(x\right)\mathrm{tan}\left(x\right)\end{array}$

$\begin{array}{l}+\cdots \\ +\frac{\left(2k-1\right)\left(2k-3\right)\left(2k-5\right)\left(2k-7\right)\cdots \left(5\right)}{2k\left(2k-2\right)\left(2k-4\right)\left(2k-6\right)\left(2k-8\right)\cdots \left(4\right)}{\mathrm{sec}}^{3}\left(x\right)\mathrm{tan}\left(x\right)\\ +\frac{\left(2k-1\right)\left(2k-3\right)\left(2k-5\right)\left(2k-7\right)\cdots \left(5\right)\left(3\right)}{2k\left(2k-2\right)\left(2k-4\right)\left(2k-6\right)\left(2k-8\right)\cdots \left(4\right)\left(2\right)}\mathrm{sec}\left(x\right)\mathrm{tan}\left(x\right)\\ +\frac{\left(2k-1\right)\left(2k-3\right)\left(2k-5\right)\left(2k-7\right)\cdots \left(5\right)\left(3\right)}{2k\left(2k-2\right)\left(2k-4\right)\left(2k-6\right)\left(2k-8\right)\cdots \left(4\right)\left(2\right)}\mathrm{ln}|\mathrm{sec}\left(x\right)+\mathrm{tan}\left(x\right)|+c\end{array}$

$\begin{array}{l}\therefore \int {\mathrm{sec}}^{2k+1}\left(x\right)\text{d}x=\frac{\underset{i=1}{\overset{k-1}{\prod }}\left(2i+1\right)}{\underset{i=1}{\overset{k}{\prod }}\left(2i\right)}\mathrm{ln}|\mathrm{sec}\left(x\right)+\mathrm{tan}\left(x\right)|+\frac{1}{2k}{\mathrm{sec}}^{2k-1}\left(x\right)\mathrm{tan}\left(x\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }\text{ }+\mathrm{tan}\left(x\right)\underset{j=1}{\overset{k-1}{\sum }}\frac{\underset{i=j}{\overset{k-1}{\prod }}\left(2i+1\right)}{\underset{i=j}{\overset{k}{\prod }}\left(2i\right)}{\mathrm{sec}}^{2i-1}\left(x\right)+c\end{array}$.

Lemma 2.5: For all $k\in N$. then:

$\int {\mathrm{csc}}^{2k+1}\left(x\right)\text{d}x=\frac{1}{2k}\left[-{\mathrm{csc}}^{2k-1}\left(x\right)\mathrm{cot}\left(x\right)+\left(2k-1\right)\int {\mathrm{csc}}^{2k-1}\left(x\right)\text{d}x\right]$

Proof: $\int {\mathrm{csc}}^{2k+1}\left(x\right)\text{d}x=\int {\mathrm{csc}}^{2k-1}\left(x\right){\mathrm{csc}}^{2}\left(x\right)\text{d}x$

Using integration by parts:

Let $u={\mathrm{csc}}^{2k-1}\left(x\right)$. then $\text{d}u=-\left(2k-1\right){\mathrm{csc}}^{2k-1}\left(x\right)\mathrm{cot}\left(x\right)\text{d}x$

$\text{d}v={\mathrm{csc}}^{2}\left(x\right)\text{d}x$. then $v=-\mathrm{cot}\left(x\right)$

$\begin{array}{l}\therefore \int {\mathrm{csc}}^{2k+1}\left(x\right)\text{d}x\\ =-{\mathrm{csc}}^{2k-1}\left(x\right)\mathrm{cot}\left(x\right)-\left(2k-1\right)\int {\mathrm{csc}}^{2k-1}\left(x\right){\mathrm{cot}}^{2}\left(x\right)\text{d}x\\ =-{\mathrm{csc}}^{2k-1}\left(x\right)\mathrm{cot}\left(x\right)-\left(2k-1\right)\int {\mathrm{csc}}^{2k-1}\left(x\right)\left({\mathrm{csc}}^{2}\left(x\right)-1\right)\text{d}x\\ =-{\mathrm{csc}}^{2k-1}\left(x\right)\mathrm{cot}\left(x\right)-\left(2k-1\right)\int {\mathrm{csc}}^{2k+1}\left(x\right)\text{d}x+\left(2k-1\right)\int {\mathrm{csc}}^{2k-1}\left(x\right)\text{d}x\end{array}$

$\therefore 2k\int {\mathrm{csc}}^{2k+1}\left(x\right)\text{d}x=-{\mathrm{csc}}^{2k-1}\left(x\right)\mathrm{cot}\left(x\right)+\left(2k-1\right)\int {\mathrm{csc}}^{2k-1}\left(x\right)\text{d}x$

$\therefore \int {\mathrm{csc}}^{2k+1}\left(x\right)\text{d}x=\frac{1}{2k}\left[-{\mathrm{csc}}^{2k-1}\left(x\right)\mathrm{tan}\left(x\right)+\left(2k-1\right)\int {\mathrm{csc}}^{2k-1}\left(x\right)\text{d}x\right].$.

Note: From clearly that: $\int {\mathrm{csc}}^{3}\left(x\right)\text{d}x=-\frac{1}{2}\left[\mathrm{csc}\left(x\right)\mathrm{cot}\left(x\right)+\mathrm{ln}|\mathrm{csc}\left(x\right)+\mathrm{cot}\left(x\right)|\right]+c$.

Theorem 2.6: For all $k\in N$. then:

$\begin{array}{l}\int {\mathrm{csc}}^{2k+1}\left(x\right)\text{d}x=-\frac{\underset{i=1}{\overset{k-1}{\prod }}\left(2i+1\right)}{\underset{i=1}{\overset{k}{\prod }}\left(2i\right)}\mathrm{ln}|\mathrm{csc}\left(x\right)+\mathrm{cot}\left(x\right)|-\frac{1}{2k}{\mathrm{csc}}^{2k-1}\left(x\right)\mathrm{cot}\left(x\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }-\mathrm{cot}\left(x\right)\underset{j=1}{\overset{k-1}{\sum }}\frac{\underset{i=j}{\overset{k-1}{\prod }}\left(2i+1\right)}{\underset{i=j}{\overset{k}{\prod }}\left(2i\right)}{\mathrm{csc}}^{2i-1}\left(x\right)+c\end{array}$

Proof: Since $\int {\mathrm{csc}}^{2k+1}\left(x\right)\text{d}x=\frac{1}{2k}\left[-{\mathrm{csc}}^{2k-1}\left(x\right)\mathrm{cot}\left(x\right)+\left(2k-1\right)\int {\mathrm{csc}}^{2k-1}\left(x\right)\text{d}x\right]$.by Lemma 2.3) Then

$\int {\mathrm{csc}}^{2k-1}\left(x\right)\text{d}x=\frac{1}{2k-2}\left[-{\mathrm{csc}}^{2k-3}\left(x\right)\mathrm{cot}\left(x\right)+\left(2k-3\right)\int {\mathrm{csc}}^{2k-3}\left(x\right)\text{d}x\right]$

and $\int {\mathrm{csc}}^{2k-3}\left(x\right)\text{d}x=\frac{1}{2k-4}\left[-{\mathrm{csc}}^{2k-5}\left(x\right)\mathrm{cot}\left(x\right)+\left(2k-5\right)\int {\mathrm{csc}}^{2k-5}\left(x\right)\text{d}x\right]$.

...

and so on

$\int {\mathrm{csc}}^{5}\left(x\right)\text{d}x=\frac{1}{4}\left[-{\mathrm{csc}}^{3}\left(x\right)\mathrm{cot}\left(x\right)+3\int {\mathrm{csc}}^{3}\left(x\right)\text{d}x\right]$

and $\int {\mathrm{csc}}^{3}\left(x\right)\text{d}x=\frac{1}{2}\left[-\mathrm{csc}\left(x\right)\mathrm{cot}\left(x\right)-\mathrm{ln}|\mathrm{csc}\left(x\right)+\mathrm{cot}\left(x\right)|\right]+c$.

$\begin{array}{l}\therefore \int {\mathrm{csc}}^{2k+1}\left(x\right)\text{d}x\\ =\frac{-1}{2k}{\mathrm{csc}}^{2k-1}\left(x\right)\mathrm{cot}\left(x\right)-\frac{2k-1}{2k\left(2k-2\right)}{\mathrm{csc}}^{2k-3}\left(x\right)\mathrm{cot}\left(x\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }-\frac{\left(2k-1\right)\left(2k-3\right)}{2k\left(2k-2\right)\left(2k-4\right)}{\mathrm{csc}}^{2k-5}\left(x\right)\mathrm{cot}\left(x\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }-\frac{\left(2k-1\right)\left(2k-3\right)\left(2k-5\right)}{2k\left(2k-2\right)\left(2k-4\right)\left(2k-6\right)}{\mathrm{csc}}^{2k-7}\left(x\right)\mathrm{cot}\left(x\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }-\frac{\left(2k-1\right)\left(2k-3\right)\left(2k-5\right)\left(2k-7\right)}{2k\left(2k-2\right)\left(2k-4\right)\left(2k-6\right)\left(2k-8\right)}{\mathrm{csc}}^{2k-9}\left(x\right)\mathrm{cot}\left(x\right)\end{array}$

$\begin{array}{l}-\cdots \\ -\frac{\left(2k-1\right)\left(2k-3\right)\left(2k-5\right)\left(2k-7\right)\cdots \left(5\right)}{2k\left(2k-2\right)\left(2k-4\right)\left(2k-6\right)\left(2k-8\right)\cdots \left(4\right)}{\mathrm{csc}}^{3}\left(x\right)\mathrm{cot}\left(x\right)\\ -\frac{\left(2k-1\right)\left(2k-3\right)\left(2k-5\right)\left(2k-7\right)\cdots \left(5\right)\left(3\right)}{2k\left(2k-2\right)\left(2k-4\right)\left(2k-6\right)\left(2k-8\right)\cdots \left(4\right)\left(2\right)}\mathrm{csc}\left(x\right)\mathrm{cot}\left(x\right)\\ -\frac{\left(2k-1\right)\left(2k-3\right)\left(2k-5\right)\left(2k-7\right)\cdots \left(5\right)\left(3\right)}{2k\left(2k-2\right)\left(2k-4\right)\left(2k-6\right)\left(2k-8\right)\cdots \left(4\right)\left(2\right)}\mathrm{ln}|\mathrm{csc}\left(x\right)+\mathrm{cot}\left(x\right)|+c\end{array}$

$\begin{array}{l}\therefore \int {\mathrm{csc}}^{2k+1}\left(x\right)\text{d}x=-\frac{\underset{i=1}{\overset{k-1}{\prod }}\left(2i+1\right)}{\underset{i=1}{\overset{k}{\prod }}\left(2i\right)}\mathrm{ln}|\mathrm{csc}\left(x\right)+\mathrm{cot}\left(x\right)|-\frac{1}{2k}{\mathrm{csc}}^{2k-1}\left(x\right)\mathrm{cot}\left(x\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }-\mathrm{cot}\left(x\right)\underset{j=1}{\overset{k-1}{\sum }}\frac{\underset{i=j}{\overset{k-1}{\prod }}\left(2i+1\right)}{\underset{i=j}{\overset{k}{\prod }}\left(2i\right)}{\mathrm{csc}}^{2i-1}\left(x\right)+c\end{array}$.

3. The Integration of Multiply Trigonometric Function Power

Theorem 3.1: For all $n\in N$. then:

1) $\int {\mathrm{sin}}^{n}\left(x\right)\mathrm{cos}\left(x\right)\text{d}x=\left\{\begin{array}{l}\frac{1}{n+1}{\mathrm{sin}}^{n+1}\left(x\right)+c,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}n\ne -1\\ \mathrm{ln}|\mathrm{sin}\left(x\right)|+c,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}n=-1\end{array}$.

2) $\int {\mathrm{cos}}^{n}\left(x\right)\mathrm{sin}\left(x\right)\text{d}x=\left\{\begin{array}{l}\frac{-1}{n+1}{\mathrm{cos}}^{n+1}\left(x\right)+c,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}n\ne -1\\ -\mathrm{ln}|\mathrm{cos}\left(x\right)|+c,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}n=-1\end{array}$.

3) $\int {\mathrm{tan}}^{n}\left(x\right){\mathrm{sec}}^{2}\left(x\right)\text{d}x=\left\{\begin{array}{l}\frac{1}{n+1}{\mathrm{tan}}^{n+1}\left(x\right)+c,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}n\ne -1\\ \mathrm{ln}|\mathrm{tan}\left(x\right)|+c,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}n=-1\end{array}$.

4) $\int {\mathrm{cot}}^{n}\left(x\right){\mathrm{csc}}^{2}\left(x\right)\text{d}x=\left\{\begin{array}{l}\frac{-1}{n+1}{\mathrm{cot}}^{n+1}\left(x\right)+c,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}n\ne -1\\ -\mathrm{ln}|\mathrm{cot}\left(x\right)|+c,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}n=-1\end{array}$.

5) $\int {\mathrm{sec}}^{n}\left(x\right)\mathrm{tan}\left(x\right)\text{d}x=\left\{\begin{array}{l}\frac{1}{n}{\mathrm{sec}}^{n}\left(x\right)+c,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}n\ne -1\\ -\mathrm{cos}\left(x\right)+c,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}n=-1\end{array}$.

6) $\int {\mathrm{csc}}^{n}\left(x\right)\mathrm{cot}\left(x\right)\text{d}x=\left\{\begin{array}{l}\frac{-1}{n}{\mathrm{csc}}^{n}\left(x\right)+c,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}n\ne -1\\ \mathrm{sin}\left(x\right)+c,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}n=-1\end{array}$.

where c is integral constant.

Proof: Directed proof. #

Theorem 3.2: For all $n,m\in N$.nd m is an odd number, then: $\int {\mathrm{sin}}^{m}\left(x\right){\mathrm{cos}}^{n}\left(x\right)\text{d}x=\underset{i=0}{\overset{k}{\sum }}\frac{{\left(-1\right)}^{i+1}}{n+1+2i}\left(\begin{array}{c}k\\ i\end{array}\right){\mathrm{cos}}^{n+1+2i}\left(x\right)+c$.

Proof: Since m is an odd number, then $\exists k\in N$.uch that $m=2k+1$.

$\begin{array}{l}\int {\mathrm{sin}}^{m}\left(x\right){\mathrm{cos}}^{n}\left(x\right)\text{d}x\\ =\int {\mathrm{sin}}^{2k+1}\left(x\right){\mathrm{cos}}^{n}\left(x\right)\text{d}x=\int {\left({\mathrm{sin}}^{2}\left(x\right)\right)}^{k}\mathrm{sin}\left(x\right){\mathrm{cos}}^{n}\left(x\right)\text{d}x\\ =\int {\left(1-{\mathrm{cos}}^{2}\left(x\right)\right)}^{k}\mathrm{sin}\left(x\right){\mathrm{cos}}^{n}\left(x\right)\text{d}x=\int \underset{i=0}{\overset{k}{\sum }}{\left(-1\right)}^{i}\left(\begin{array}{c}k\\ i\end{array}\right){\mathrm{cos}}^{2i}\left(x\right)\mathrm{sin}\left(x\right){\mathrm{cos}}^{n}\left(x\right)\text{d}x\end{array}$

$=\underset{i=0}{\overset{k}{\sum }}{\left(-1\right)}^{i}\left(\begin{array}{c}k\\ i\end{array}\right)\int {\mathrm{cos}}^{n+2i}\left(x\right)\mathrm{sin}\left(x\right)dx=\underset{i=0}{\overset{k}{\sum }}\frac{{\left(-1\right)}^{i+1}}{n+1+2i}\left(\begin{array}{c}k\\ i\end{array}\right){\mathrm{cos}}^{n+1+2i}\left(x\right)+c$.

Theorem 3.3: For all $n,m\in N$.nd m is an even number, then:

$\begin{array}{l}\int {\mathrm{sin}}^{m}\left(x\right){\mathrm{cos}}^{n}\left(x\right)\text{d}x\\ =\left\{\begin{array}{l}\underset{i=0}{\overset{k}{\sum }}{\left(-1\right)}^{i}\left(\begin{array}{c}k\\ i\end{array}\right)\left[\frac{1}{{2}^{\alpha }}\left(\begin{array}{c}\alpha \\ \frac{\alpha }{2}\end{array}\right)x+\frac{1}{{2}^{\alpha -1}}\underset{j=0}{\overset{\frac{\alpha }{2}-1}{\sum }}\frac{\left(\begin{array}{c}\alpha \\ j\end{array}\right)}{\alpha -2j}\mathrm{sin}\left(\alpha -2j\right)x\right]+c\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{if}\text{\hspace{0.17em}}\alpha =2i+n,\text{\hspace{0.17em}}\text{\hspace{0.17em}}n\text{\hspace{0.17em}}\text{is}\text{\hspace{0.17em}}\text{even}\\ \underset{i=0}{\overset{k}{\sum }}{\left(-1\right)}^{i}\left(\begin{array}{c}k\\ i\end{array}\right)\left[\frac{1}{{2}^{\alpha -1}}\underset{j=1}{\overset{\frac{\alpha +1}{2}}{\sum }}\frac{\left(\begin{array}{c}\alpha -1\\ j-1\end{array}\right)}{\alpha -2j+1}\mathrm{sin}\left(\alpha -2j+1\right)x\right]+c\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{if}\text{\hspace{0.17em}}\alpha =2i+n,\text{\hspace{0.17em}}\text{\hspace{0.17em}}n\text{\hspace{0.17em}}\text{is}\text{\hspace{0.17em}}\text{odd}\end{array}\end{array}$.span class="bracketMark">(3.3)

Proof: Since m is an even number, then $\exists k\in N$.uch that $m=2k$.

$\begin{array}{l}\int {\mathrm{sin}}^{m}\left(x\right){\mathrm{cos}}^{n}\left(x\right)\text{d}x\\ =\int {\mathrm{sin}}^{2k}\left(x\right){\mathrm{cos}}^{n}\left(x\right)\text{d}x\\ =\int {\left({\mathrm{sin}}^{2}\left(x\right)\right)}^{k}{\mathrm{cos}}^{n}\left(x\right)\text{d}x\\ =\int {\left(1-{\mathrm{cos}}^{2}\left(x\right)\right)}^{k}{\mathrm{cos}}^{n}\left(x\right)\text{d}x\\ =\int \underset{i=0}{\overset{k}{\sum }}{\left(-1\right)}^{i}\left(\begin{array}{c}k\\ i\end{array}\right){\mathrm{cos}}^{2i+n}\left(x\right)\text{d}x\end{array}$

By Theorem 2.1, we have (3.3).

Theorem 3.4: For all $n,m\in N$.nd m is an odd number, then:

$\int {\mathrm{tan}}^{m}\left(x\right){\mathrm{sec}}^{n}\left(x\right)\text{d}x=\underset{i=0}{\overset{k}{\sum }}\frac{{\left(-1\right)}^{k+i}}{n+2i}\left(\begin{array}{l}k\\ i\end{array}\right){\mathrm{sec}}^{n+2i}\left(x\right)+c$.

Proof: Since m is an odd number, then $\exists k\in N$.uch that $m=2k+1$.

$\begin{array}{l}\int {\mathrm{tan}}^{m}\left(x\right){\mathrm{sec}}^{n}\left(x\right)\text{d}x\\ =\int {\mathrm{tan}}^{2k+1}\left(x\right){\mathrm{sec}}^{n}\left(x\right)\text{d}x=\int {\left({\mathrm{tan}}^{2}\left(x\right)\right)}^{k}\mathrm{tan}\left(x\right){\mathrm{sec}}^{n}\left(x\right)\text{d}x\\ =\int {\left({\mathrm{sec}}^{2}\left(x\right)-1\right)}^{k}\mathrm{tan}\left(x\right){\mathrm{sec}}^{n}\left(x\right)\text{d}x\end{array}$

$\begin{array}{l}=\int {\left(-1\right)}^{k}{\left(1-{\mathrm{sec}}^{2}\left(x\right)\right)}^{k}\mathrm{tan}\left(x\right){\mathrm{sec}}^{n}\left(x\right)\text{d}x\\ =\int {\left(-1\right)}^{k}\underset{i=0}{\overset{k}{\sum }}{\left(-1\right)}^{i}\left(\begin{array}{c}k\\ i\end{array}\right){\mathrm{sec}}^{2i}\left(x\right)\mathrm{tan}\left(x\right){\mathrm{sec}}^{n}\left(x\right)\text{d}x\\ =\underset{i=0}{\overset{k}{\sum }}{\left(-1\right)}^{k+i}\left(\begin{array}{c}k\\ i\end{array}\right)\int {\mathrm{sec}}^{n+2i-1}\left(x\right)\mathrm{sec}\left(x\right)\mathrm{tan}\left(x\right)\text{d}x\\ =\underset{i=0}{\overset{k}{\sum }}\frac{{\left(-1\right)}^{k+i}}{n+2i}\left(\begin{array}{c}k\\ i\end{array}\right){\mathrm{sec}}^{n+2i}\left(x\right)+c\end{array}$.

Theorem 3.5: For all $n,m\in N$.nd m is an even number, then: $\int {\mathrm{tan}}^{m}\left(x\right)\text{\hspace{0.17em}}{\mathrm{sec}}^{n}\left(x\right)\text{\hspace{0.17em}}dx=$

$\begin{array}{l}\int {\mathrm{tan}}^{m}\left(x\right){\mathrm{sec}}^{n}\left(x\right)\text{d}x\\ =\left\{\begin{array}{l}\underset{i=0}{\overset{k}{\sum }}{\left(-1\right)}^{k+i}\left(\begin{array}{c}k\\ i\end{array}\right)\left[\underset{j=0}{\overset{\frac{\alpha }{2}-1}{\sum }}\frac{\left(\begin{array}{c}\frac{\alpha }{2}-1\\ j\end{array}\right)}{2j+1}{\mathrm{tan}}^{2j+1}\left(x\right)\right]+c,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{if}\text{\hspace{0.17em}}\alpha =2i+n,\text{\hspace{0.17em}}\text{\hspace{0.17em}}n\text{\hspace{0.17em}}\text{is}\text{\hspace{0.17em}}\text{even}\\ \underset{i=0}{\overset{k}{\sum }}{\left(-1\right)}^{k+i}\left(\begin{array}{c}k\\ i\end{array}\right)\left[\frac{1}{\alpha -1}{\mathrm{sec}}^{\alpha -2}\left(x\right)\mathrm{tan}\left(x\right)+\frac{\underset{m=1}{\overset{\frac{\alpha -3}{2}}{\prod }}\left(2m+1\right)}{\underset{m=1}{\overset{\frac{\alpha -1}{2}}{\prod }}\left(2m\right)}\mathrm{ln}|\mathrm{sec}\left(x\right)+\mathrm{tan}\left(x\right)|\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+\mathrm{tan}\left(x\right)\underset{j=1}{\overset{\frac{\alpha -3}{2}}{\sum }}\frac{\underset{m=j}{\overset{\frac{\alpha -3}{2}}{\prod }}\left(2m+1\right)}{\underset{m=j}{\overset{\frac{\alpha -1}{2}}{\prod }}\left(2m\right)}{\mathrm{sec}}^{2m-1}\left(x\right)+c\right],\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{if}\text{\hspace{0.17em}}\alpha =2i+n,\text{\hspace{0.17em}}\text{\hspace{0.17em}}n\text{\hspace{0.17em}}\text{is}\text{\hspace{0.17em}}\text{odd}\end{array}\end{array}$.span class="bracketMark">(3.5)

Proof: Since m is an even number, then $\exists k\in N$.uch that $m=2k$.

$\begin{array}{l}\int {\mathrm{tan}}^{m}\left(x\right){\mathrm{sec}}^{n}\left(x\right)\text{d}x\\ =\int {\mathrm{tan}}^{2k}\left(x\right){\mathrm{sec}}^{n}\left(x\right)\text{d}x=\int {\left({\mathrm{tan}}^{2}\left(x\right)\right)}^{k}{\mathrm{sec}}^{n}\left(x\right)\text{d}x\end{array}$

$\begin{array}{l}=\int {\left({\mathrm{sec}}^{2}\left(x\right)-1\right)}^{k}{\mathrm{sec}}^{n}\left(x\right)\text{d}x=\int {\left(-1\right)}^{k}{\left(1-{\mathrm{sec}}^{2}\left(x\right)\right)}^{k}{\mathrm{sec}}^{n}\left(x\right)\text{d}x\\ =\int \underset{i=0}{\overset{k}{\sum }}{\left(-1\right)}^{k+i}\left(\begin{array}{c}k\\ i\end{array}\right){\mathrm{sec}}^{2i+n}\left(x\right)\text{d}x\end{array}$

By Theorem 2.2, we have (3.5). #

Theorem 3.6: For all $n,m\in N$.nd m is an odd number, then: $\int {\mathrm{cot}}^{m}\left(x\right){\mathrm{csc}}^{n}\left(x\right)\text{d}x=\underset{i=0}{\overset{k}{\sum }}\frac{{\left(-1\right)}^{k+i+1}}{n+2i}\left(\begin{array}{c}k\\ i\end{array}\right){\mathrm{csc}}^{n+2i}\left(x\right)+c$.

Proof: Since m is an odd number, then $\exists k\in N$.uch that $m=2k+1$.

$\begin{array}{l}\int {\mathrm{cot}}^{m}\left(x\right){\mathrm{csc}}^{n}\left(x\right)\text{d}x\\ =\int {\mathrm{cot}}^{2k+1}\left(x\right){\mathrm{csc}}^{n}\left(x\right)\text{d}x=\int {\mathrm{cot}}^{2k}\left(x\right)\mathrm{cot}\left(x\right){\mathrm{csc}}^{n}\left(x\right)\text{d}x\\ =\int {\left({\mathrm{cot}}^{2}\left(x\right)\right)}^{k}\mathrm{cot}\left(x\right){\mathrm{csc}}^{n}\left(x\right)\text{d}x=\int {\left({\mathrm{csc}}^{2}\left(x\right)-1\right)}^{k}\mathrm{cot}\left(x\right){\mathrm{csc}}^{n}\left(x\right)\text{d}x\end{array}$ $\begin{array}{l}=\int {\left(-1\right)}^{k}{\left(1-{\mathrm{csc}}^{2}\left(x\right)\right)}^{k}\mathrm{cot}\left(x\right){\mathrm{csc}}^{n}\left(x\right)\text{d}x\\ =\int {\left(-1\right)}^{k}\underset{i=0}{\overset{k}{\sum }}{\left(-1\right)}^{i}\left(\begin{array}{c}k\\ i\end{array}\right){\mathrm{csc}}^{2i}\left(x\right)\mathrm{cot}\left(x\right){\mathrm{csc}}^{n}\left(x\right)\text{d}x\\ =\underset{i=0}{\overset{k}{\sum }}-{\left(-1\right)}^{k+i}\left(\begin{array}{c}k\\ i\end{array}\right)\int {\mathrm{csc}}^{n+2i-1}\left(x\right)\left(-\mathrm{csc}\left(x\right)\right)\mathrm{cot}\left(x\right)\text{d}x\\ =\underset{i=0}{\overset{k}{\sum }}\frac{{\left(-1\right)}^{k+i+1}}{n+2i}\left(\begin{array}{c}k\\ i\end{array}\right){\mathrm{csc}}^{n+2i}\left(x\right)+c\end{array}$.

Theorem 3.7: For all $n,m\in N$.nd m is an even number, then: $\int {\mathrm{cot}}^{m}\left(x\right)\text{\hspace{0.17em}}{\mathrm{csc}}^{n}\left(x\right)\text{\hspace{0.17em}}dx=$ $\begin{array}{l}\int {\mathrm{cot}}^{m}\left(x\right){\mathrm{csc}}^{n}\left(x\right)\text{d}x\\ =\left\{\begin{array}{l}-\underset{i=0}{\overset{k}{\sum }}{\left(-1\right)}^{k+i}\left(\begin{array}{c}k\\ i\end{array}\right)\left[\underset{j=0}{\overset{\frac{\alpha }{2}-1}{\sum }}\frac{\left(\begin{array}{c}\frac{\alpha }{2}-1\\ j\end{array}\right)}{2j+1}{\mathrm{cot}}^{2j+1}\left(x\right)\right]+c,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{if}\text{\hspace{0.17em}}\alpha =2i+n,\text{\hspace{0.17em}}\text{\hspace{0.17em}}n\text{\hspace{0.17em}}\text{is}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{even}\\ -\underset{i=0}{\overset{k}{\sum }}{\left(-1\right)}^{k+i}\left(\begin{array}{c}k\\ i\end{array}\right)\left[\frac{1}{\alpha -1}{\mathrm{csc}}^{\alpha -2}\left(x\right)\mathrm{cot}\left(x\right)+\frac{\underset{m=1}{\overset{\frac{\alpha -3}{2}}{\prod }}\left(2m+1\right)}{\underset{m=1}{\overset{\frac{\alpha -1}{2}}{\prod }}\left(2m\right)}\mathrm{ln}|\mathrm{csc}\left(x\right)+\mathrm{cot}\left(x\right)|\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+\mathrm{cot}\left(x\right)\underset{j=1}{\overset{\frac{\alpha -3}{2}}{\sum }}\frac{\underset{m=j}{\overset{\frac{\alpha -3}{2}}{\prod }}\left(2m+1\right)}{\underset{m=j}{\overset{\frac{\alpha -1}{2}}{\prod }}\left(2m\right)}{\mathrm{csc}}^{2m-1}\left(x\right)\right]+c,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{if}\text{\hspace{0.17em}}\alpha =2i+n,\text{\hspace{0.17em}}\text{\hspace{0.17em}}n\text{\hspace{0.17em}}\text{is}\text{\hspace{0.17em}}\text{odd}\end{array}\end{array}$.span class="bracketMark">(3.7)

Proof: Since m is an even number, then $\exists k\in N$.uch that $m=2k$

$\begin{array}{l}\int {\mathrm{cot}}^{m}\left(x\right){\mathrm{csc}}^{n}\left(x\right)\text{d}x\\ =\int {\mathrm{cot}}^{2k}\left(x\right){\mathrm{csc}}^{n}\left(x\right)\text{d}x\\ =\int {\left({\mathrm{cot}}^{2}\left(x\right)\right)}^{k}{\mathrm{csc}}^{n}\left(x\right)\text{d}x\\ =\int {\left({\mathrm{csc}}^{2}\left(x\right)-1\right)}^{k}{\mathrm{csc}}^{n}\left(x\right)\text{d}x\end{array}$

$\begin{array}{l}=\int {\left(-1\right)}^{k}\underset{i=0}{\overset{k}{\sum }}{\left(-1\right)}^{i}\left(\begin{array}{c}k\\ i\end{array}\right){\mathrm{csc}}^{2i+n}\left(x\right)\text{d}x\\ =\int \underset{i=0}{\overset{k}{\sum }}{\left(-1\right)}^{k+i}\left(\begin{array}{c}k\\ i\end{array}\right){\mathrm{csc}}^{2i+n}\left(x\right)\text{d}x\\ =\underset{i=0}{\overset{k}{\sum }}{\left(-1\right)}^{k+i}\left(\begin{array}{c}k\\ i\end{array}\right)\int {\mathrm{csc}}^{2i+n}\left(x\right)\text{d}x\end{array}$

Using Theorems 2.2 and 2.5, we have (3.7). #

Remark: There are many integration formulas, we can be finding it by previous results which obtained it from this paper. For example:

1) $\int {\mathrm{sin}}^{m}\left(x\right)f\left(x\right)\text{d}x$. $\int {\mathrm{cos}}^{m}\left(x\right)f\left(x\right)\text{d}x$.

where $f\left(x\right)\in \left\{{\mathrm{tan}}^{n}\left(x\right),{\mathrm{cot}}^{n}\left(x\right),{\mathrm{sec}}^{n}\left(x\right),{\mathrm{csc}}^{n}\left(x\right)\right\}$. $\forall n,m\in N$.

2) $\int {\mathrm{tan}}^{m}\left(x\right)g\left(x\right)\text{d}x$. where $g\left(x\right)\in \left\{{\mathrm{cot}}^{n}\left(x\right),{\mathrm{sec}}^{n}\left(x\right),{\mathrm{csc}}^{n}\left(x\right)\right\}$. $\forall n,m\in N$.

3) $\int {\mathrm{cot}}^{m}\left(x\right){\mathrm{sec}}^{n}\left(x\right)\text{d}x$.

4. The Hyperbolic Functions

The combinations of the exponential functions $\mathrm{exp}\left(x\right)$.nd $\mathrm{exp}\left(-x\right)$.alled hyperbolic functions. These functions which arise in various engineering applications, have many properties in common with the trigonometric functions. The hyperbolic functions have resulted from vibratory motions inside elastic solid and more general in many problems where mechanical energy is gradually absorbed by a surrounding medium. There are importance applications on hyperbolic functions and power integration of hyperbolic functions in physics, mathematics transformations and numerical analysis.

Remark: The proofs for this following result are similarly of the results for previous sections.

4.1. The Hyperbolic Functions Power

Theorem 4.1.1: For all $k\in N$. then:

1) ${\mathrm{sinh}}^{2k-1}\left(x\right)=\frac{1}{{2}^{2k-2}}\underset{n=0}{\overset{k-1}{\sum }}{\left(-1\right)}^{n}\left(\begin{array}{c}2k-1\\ n\end{array}\right)\mathrm{sinh}\left(2k-1-2n\right)x$.

2) ${\mathrm{sinh}}^{2k}\left(x\right)=\frac{{\left(-1\right)}^{k}}{{2}^{2k}}\left(\begin{array}{c}2k\\ k\end{array}\right)+\frac{1}{{2}^{2k-1}}\underset{n=0}{\overset{k-1}{\sum }}{\left(-1\right)}^{n}\left(\begin{array}{c}2k\\ n\end{array}\right)\mathrm{cosh}\left(2k-2n\right)x$.

3) ${\mathrm{cosh}}^{2k-1}\left(x\right)=\frac{1}{{2}^{2k-2}}\underset{n=0}{\overset{k-1}{\sum }}\left(\begin{array}{c}2k-1\\ n\end{array}\right)\mathrm{cosh}\left(2k-1-2n\right)x$.

4) ${\mathrm{cosh}}^{2k}\left(x\right)=\frac{1}{{2}^{2k}}\left(\begin{array}{c}2k\\ k\end{array}\right)+\frac{1}{{2}^{2k-1}}\underset{n=0}{\overset{k-1}{\sum }}\left(\begin{array}{c}2k\\ n\end{array}\right)\mathrm{cosh}\left(2k-2n\right)x$.

5) ${\mathrm{tanh}}^{2k}\left(x\right)=1-{\text{sech}}^{2}\left(x\right)\underset{n=1}{\overset{k}{\sum }}{\mathrm{tanh}}^{2k-2n}\left(x\right)$.

6) ${\mathrm{tanh}}^{2k+1}\left(x\right)=\mathrm{tanh}\left(x\right)-{\text{sech}}^{2}\left(x\right)\underset{n=1}{\overset{k}{\sum }}{\mathrm{tanh}}^{2k-2n+1}\left(x\right)$.

7) ${\mathrm{coth}}^{2k}\left(x\right)=1+{\text{csch}}^{2}\left(x\right)\underset{n=1}{\overset{k}{\sum }}{\mathrm{coth}}^{2k-2n}\left(x\right)$.

8) ${\mathrm{coth}}^{2k+1}\left(x\right)=\mathrm{coth}\left(x\right)+{\text{csch}}^{2}\left(x\right)\underset{n=1}{\overset{k}{\sum }}{\mathrm{coth}}^{2k-2n+1}\left(x\right)$.

9) ${\text{sech}}^{2k}\left(x\right)={\text{sech}}^{2}\left(x\right)\underset{n=0}{\overset{k-1}{\sum }}{\left(-1\right)}^{n}\left(\begin{array}{c}k-1\\ n\end{array}\right){\mathrm{tanh}}^{2n}\left(x\right)$.

10) ${\text{csch}}^{2k}\left(x\right)={\text{csch}}^{2}\left(x\right)\underset{n=0}{\overset{k-1}{\sum }}{\left(-1\right)}^{k+n-1}\left(\begin{array}{c}k-1\\ n\end{array}\right){\mathrm{coth}}^{2n}\left(x\right)$.

4.2. The Integration of Hyperbolic Functions Power

Theorem 4.2.1: For all $k\in N$. then:

1) $\int {\mathrm{sinh}}^{2k-1}\left(x\right)=\frac{1}{{2}^{2k-2}}\underset{n=0}{\overset{k-1}{\sum }}\frac{{\left(-1\right)}^{n}}{2k-1-2n}\left(\begin{array}{c}2k-1\\ n\end{array}\right)\mathrm{cosh}\left(2k-1-2n\right)x+c$

2) $\int {\mathrm{sinh}}^{2k}\left(x\right)=\frac{{\left(-1\right)}^{k}}{{2}^{2k}}\left(\begin{array}{c}2k\\ n\end{array}\right)x+\frac{1}{{2}^{2k-1}}\underset{n=0}{\overset{k-1}{\sum }}\frac{{\left(-1\right)}^{n}}{2k-2n}\left(\begin{array}{c}2k\\ n\end{array}\right)\mathrm{sinh}\left(2k-2n\right)x+c$

3) $\int {\mathrm{cosh}}^{2k-1}\left(x\right)=\frac{1}{{2}^{2k-2}}\underset{n=0}{\overset{k-1}{\sum }}\frac{1}{2k-1-2n}\left(\begin{array}{c}2k-1\\ n\end{array}\right)\mathrm{sinh}\left(2k-1-2n\right)x+c$

4) $\int {\mathrm{cosh}}^{2k}\left(x\right)=\frac{1}{{2}^{2k}}\left(\begin{array}{c}2k\\ k\end{array}\right)x+\frac{1}{{2}^{2k-1}}\underset{n=0}{\overset{k-1}{\sum }}\frac{1}{2k-2n}\left(\begin{array}{c}2k\\ n\end{array}\right)\mathrm{sinh}\left(2k-2n\right)x+c$

5) $\int {\mathrm{tanh}}^{2k}\left(x\right)=x-\underset{n=1}{\overset{k}{\sum }}\frac{1}{2k+1-2n}{\mathrm{tanh}}^{2k+1-2n}\left(x\right)+c$.

6) $\int {\mathrm{tanh}}^{2k+1}\left(x\right)=\mathrm{ln}|\mathrm{cosh}\left(x\right)|-\underset{n=1}{\overset{k}{\sum }}\frac{1}{2k+2-2n}{\mathrm{tanh}}^{2k+2-2n}\left(x\right)+c$.

7) $\int {\mathrm{coth}}^{2k}\left(x\right)=x-\underset{n=1}{\overset{k}{\sum }}\frac{1}{2k+1-2n}{\mathrm{coth}}^{2k+1-2n}\left(x\right)+c$.

8) $\int {\mathrm{coth}}^{2k+1}\left(x\right)=\mathrm{ln}|\mathrm{sinh}\left(x\right)|-\underset{n=1}{\overset{k}{\sum }}\frac{1}{2k+2-2n}{\mathrm{coth}}^{2k-2n+2}\left(x\right)+c$.

9) $\int {\text{sech}}^{2k}\left(x\right)=\underset{n=0}{\overset{k-1}{\sum }}\frac{{\left(-1\right)}^{n}}{2n+1}\left(\begin{array}{c}k-1\\ n\end{array}\right){\mathrm{tanh}}^{2n+1}\left(x\right)+c$.

10) $\int {\text{csch}}^{2k}\left(x\right)=\underset{n=0}{\overset{k-1}{\sum }}\frac{{\left(-1\right)}^{k+n}}{2n+1}\left(\begin{array}{c}k-1\\ n\end{array}\right){\mathrm{coth}}^{2n+1}\left(x\right)+c$.

Lemma 4.2.2: For all $k\in N$. then:

1) $\int {\text{sech}}^{2k+1}\left(x\right)\text{d}x=\frac{1}{2k}\left[{\text{sech}}^{2k-1}\left(x\right)\mathrm{tanh}\left(x\right)+\left(2k-1\right)\int {\mathrm{sec}}^{2k-1}\left(x\right)\text{d}x\right]$

2) $\int {\text{csch}}^{2k+1}\left(x\right)\text{d}x=-\frac{1}{2k}\left[{\text{csch}}^{2k-1}\left(x\right)\mathrm{coth}\left(x\right)+\left(2k-1\right)\int {\text{csch}}^{2k-1}\left(x\right)\text{d}x\right]$

Theorem 4.2.3: For all $k\in N$. then:

1) $\begin{array}{l}\int {\mathrm{sec}}^{2k+1}\left(x\right)\text{d}x\\ =\frac{\underset{i=1}{\overset{k-1}{\prod }}\left(2i+1\right)}{\underset{i=1}{\overset{k}{\prod }}\left(2i\right)}{\mathrm{tan}}^{-1}{\text{e}}^{x}+\frac{1}{2k}{\text{sech}}^{2k-1}\left(x\right)\mathrm{tanh}\left(x\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }+\mathrm{tanh}\left(x\right)\underset{j=1}{\overset{k-1}{\sum }}\frac{\underset{i=j}{\overset{k-1}{\prod }}\left(2i+1\right)}{\underset{i=j}{\overset{k}{\prod }}\left(2i\right)}{\text{sech}}^{2i-1}\left(x\right)+c\end{array}$

2) $\begin{array}{l}\int {\text{csch}}^{2k+1}\left(x\right)\text{d}x\\ ={\left(-1\right)}^{n+1}\frac{\underset{i=1}{\overset{k-1}{\prod }}\left(2i+1\right)}{\underset{i=1}{\overset{k}{\prod }}\left(2i\right)}\mathrm{ln}|\text{csch}\left(x\right)+\mathrm{coth}\left(x\right)|\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }-\frac{1}{2k}{\text{csch}}^{2k-1}\left(x\right)\mathrm{coth}\left(x\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+\mathrm{coth}\left(x\right)\underset{j=1}{\overset{k-1}{\sum }}\frac{\underset{i=j}{\overset{k-1}{\prod }}\left(2i+1\right)}{\underset{i=j}{\overset{k}{\prod }}\left(2i\right)}{\text{csch}}^{2i-1}\left(x\right)+c\end{array}$

4.3. The Integration of Multiply Trigonometric Function Power

Theorem 4.3.1: For all $n\in N$. then:

1) $\int {\mathrm{sinh}}^{n}\left(x\right)\mathrm{cosh}\left(x\right)\text{d}x=\left\{\begin{array}{l}\frac{1}{n+1}{\mathrm{sinh}}^{n+1}\left(x\right)+c,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}n\ne -1\\ \mathrm{ln}|\mathrm{sinh}\left(x\right)|+c,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}n=-1\end{array}$.

2) $\int {\mathrm{cosh}}^{n}\left(x\right)\mathrm{sinh}\left(x\right)\text{d}x=\left\{\begin{array}{l}\frac{1}{n+1}{\mathrm{cosh}}^{n+1}\left(x\right)+c,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}n\ne -1\\ \mathrm{ln}\left(\mathrm{cosh}\left(x\right)\right)+c,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}n=-1\end{array}$.

3) $\int {\mathrm{tanh}}^{n}\left(x\right){\text{sech}}^{2}\left(x\right)\text{d}x=\left\{\begin{array}{l}\frac{1}{n+1}{\mathrm{tanh}}^{n+1}\left(x\right)+c,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}n\ne -1\\ \mathrm{ln}|\mathrm{tanh}\left(x\right)|+c,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}n=-1\end{array}$.

4) $\int {\mathrm{coth}}^{n}\left(x\right){\text{csch}}^{2}\left(x\right)\text{d}x=\left\{\begin{array}{l}\frac{-1}{n+1}{\mathrm{coth}}^{n+1}\left(x\right)+c,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}n\ne -1\\ -\mathrm{ln}|\mathrm{coth}\left(x\right)|+c,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}n=-1\end{array}$.

5) $\int {\text{sech}}^{n}\left(x\right)\mathrm{tanh}\left(x\right)\text{d}x=\left\{\begin{array}{l}\frac{1}{n}{\text{sech}}^{n}\left(x\right)+c,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}n\ne -1\\ \mathrm{cosh}\left(x\right)+c,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}n=-1\end{array}$.

6) $\int {\text{csch}}^{n}\left(x\right)\mathrm{coth}\left(x\right)\text{d}x=\left\{\begin{array}{l}\frac{-1}{n}{\text{csch}}^{n}\left(x\right)+c,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}n\ne -1\\ \mathrm{sinh}\left(x\right)+c,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}n=-1\end{array}$.

where c is integral constant.

Theorem 4.3.2: For all $n,m\in N$.nd m is an odd number, then:

1) $\begin{array}{l}\int {\mathrm{sinh}}^{m}\left(x\right){\mathrm{cosh}}^{n}\left(x\right)\text{d}x\\ =\underset{i=0}{\overset{k}{\sum }}\frac{{\left(-1\right)}^{k+i}}{n+2i+1}\left(\begin{array}{l}k\\ i\end{array}\right){\mathrm{cosh}}^{n+2i+1}\left(x\right)+c\end{array}$.

2) $\begin{array}{l}\int {\mathrm{tanh}}^{m}\left(x\right){\text{sech}}^{n}\left(x\right)\text{d}x\\ =\underset{i=0}{\overset{k}{\sum }}\frac{{\left(-1\right)}^{i+1}}{n+2i}\left(\begin{array}{l}k\\ i\end{array}\right){\text{sech}}^{n+2i}\left(x\right)+c\end{array}$.

3) $\begin{array}{l}\int {\mathrm{coth}}^{m}\left(x\right){\text{csch}}^{n}\left(x\right)\text{d}x\\ =-\underset{i=0}{\overset{k}{\sum }}\frac{1}{n+2i}\left(\begin{array}{c}k\\ i\end{array}\right){\text{csch}}^{n+2i}\left(x\right)+c\end{array}$.

Theorem 4.3.3: For all $n,m\in N$.nd m is an even number, then:

1) $\begin{array}{l}\int {\mathrm{sinh}}^{m}\left(x\right){\mathrm{cosh}}^{n}\left(x\right)\text{d}x\\ =\left\{\begin{array}{l}\underset{i=0}{\overset{k}{\sum }}{\left(-1\right)}^{k+i}\left(\begin{array}{c}k\\ i\end{array}\right)\left[\frac{1}{{2}^{\alpha }}\left(\begin{array}{c}\alpha \\ \frac{\alpha }{2}\end{array}\right)x+\frac{1}{{2}^{\alpha -1}}\underset{j=0}{\overset{\frac{\alpha }{2}-1}{\sum }}\frac{\left(\begin{array}{c}\alpha \\ j\end{array}\right)}{\alpha -2j}\mathrm{sinh}\left(\alpha -2j\right)x\right]+c\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{if}\text{\hspace{0.17em}}\alpha =2i+n,\text{\hspace{0.17em}}\text{\hspace{0.17em}}n\text{\hspace{0.17em}}\text{is}\text{\hspace{0.17em}}\text{even}\\ \underset{i=0}{\overset{k}{\sum }}{\left(-1\right)}^{k+i}\left(\begin{array}{c}k\\ i\end{array}\right)\left[\frac{1}{{2}^{\alpha -1}}\underset{j=0}{\overset{\frac{\alpha -1}{2}}{\sum }}\frac{\left(\begin{array}{c}\alpha \\ j-1\end{array}\right)}{\alpha -2j}\mathrm{sinh}\left(\alpha -2j\right)x\right]+c\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{if}\text{\hspace{0.17em}}\alpha =2i+n,\text{\hspace{0.17em}}\text{\hspace{0.17em}}n\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{is}\text{\hspace{0.17em}}\text{odd}\end{array}\end{array}$

2) $\begin{array}{l}\int {\mathrm{tanh}}^{m}\left(x\right){\text{sech}}^{n}\left(x\right)\text{d}x\\ =\left\{\begin{array}{l}\underset{i=0}{\overset{k}{\sum }}{\left(-1\right)}^{i}\left(\begin{array}{c}k\\ i\end{array}\right)\left[\underset{j=0}{\overset{\frac{\alpha }{2}-1}{\sum }}\frac{\left(\begin{array}{c}\frac{\alpha }{2}-1\\ j\end{array}\right)}{2j+1}{\mathrm{tanh}}^{2j+1}\left(x\right)\right]+c,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{if}\text{\hspace{0.17em}}\alpha =2i+n,\text{\hspace{0.17em}}\text{\hspace{0.17em}}n\text{\hspace{0.17em}}\text{is}\text{\hspace{0.17em}}\text{even}\\ \underset{i=0}{\overset{k}{\sum }}{\left(-1\right)}^{i}\left(\begin{array}{c}k\\ i\end{array}\right)\left[\frac{1}{\alpha -1}{\text{sech}}^{\alpha -1}\left(x\right)\mathrm{tanh}\left(x\right)+\frac{\underset{m=1}{\overset{\frac{\alpha -3}{2}}{\prod }}\left(2m+1\right)}{\underset{m=1}{\overset{\frac{\alpha -1}{2}}{\prod }}\left(2m\right)}{\mathrm{tan}}^{-1}{\text{e}}^{x}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+\mathrm{tanh}\left(x\right)\underset{j=1}{\overset{\frac{\alpha -3}{2}}{\sum }}\frac{\underset{m=j}{\overset{\frac{\alpha -3}{2}}{\prod }}\left(2m+1\right)}{\underset{m=j}{\overset{\frac{\alpha -1}{2}}{\prod }}\left(2m\right)}{\text{sech}}^{2m-1}\left(x\right)+c\right],\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{if}\text{\hspace{0.17em}}\alpha =2i+n,\text{\hspace{0.17em}}\text{\hspace{0.17em}}n\text{\hspace{0.17em}}\text{is}\text{\hspace{0.17em}}\text{odd}\end{array}\end{array}$

3) $\begin{array}{l}\int {\mathrm{coth}}^{m}\left(x\right){\text{csch}}^{n}\left(x\right)\text{d}x\\ =\left\{\begin{array}{l}\underset{i=0}{\overset{k}{\sum }}\left(\begin{array}{c}k\\ i\end{array}\right)\left[\underset{j=0}{\overset{\frac{\alpha }{2}-1}{\sum }}\frac{\left(\begin{array}{c}\frac{\alpha }{2}-1\\ j\end{array}\right)}{2j+1}{\mathrm{coth}}^{2j+1}\left(x\right)\right]+c,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{if}\text{\hspace{0.17em}}\alpha =2i+n,\text{\hspace{0.17em}}\text{\hspace{0.17em}}n\text{\hspace{0.17em}}\text{is}\text{\hspace{0.17em}}\text{even}\\ \underset{i=0}{\overset{k}{\sum }}\left(\begin{array}{c}k\\ i\end{array}\right)\left[\frac{-1}{\alpha -1}{\text{csch}}^{\alpha -1}\left(x\right)\mathrm{coth}\left(x\right)+{\left(-1\right)}^{\frac{\alpha +1}{2}}\frac{\underset{m=1}{\overset{\frac{\alpha -3}{2}}{\prod }}\left(2m+1\right)}{\underset{m=1}{\overset{\frac{\alpha -1}{2}}{\prod }}\left(2m\right)}\mathrm{ln}|\text{csch}\left(x\right)+\mathrm{coth}\left(x\right)|\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+\mathrm{coth}\left(x\right)\underset{j=1}{\overset{\frac{\alpha -3}{2}}{\sum }}{\left(-1\right)}^{\frac{\alpha +1}{2}+j}\frac{\underset{m=j}{\overset{\frac{\alpha -3}{2}}{\prod }}\left(2m+1\right)}{\underset{m=j}{\overset{\frac{\alpha -1}{2}}{\prod }}\left(2m\right)}{\text{csch}}^{2m-1}\left(x\right)\right]+c,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{if}\text{\hspace{0.17em}}\alpha =2i+n,\text{\hspace{0.17em}}\text{\hspace{0.17em}}n\text{\hspace{0.17em}}\text{is}\text{\hspace{0.17em}}\text{odd}\end{array}\end{array}$

Remark

There are many integration formulas, we can find it by previous results which obtained it from this paper. For example:

1) $\int {\mathrm{sinh}}^{m}\left(x\right)f\left(x\right)\text{d}x$. $\int {\mathrm{cosh}}^{m}\left(x\right)f\left(x\right)\text{d}x$.

where $f\left(x\right)\in \left\{{\mathrm{tanh}}^{n}\left(x\right),{\mathrm{coth}}^{n}\left(x\right),{\text{sech}}^{n}\left(x\right),{\text{csch}}^{n}\left(x\right)\right\}$. $\forall n,m\in N$.

2) $\int {\mathrm{tanh}}^{m}\left(x\right)g\left(x\right)\text{d}x$. where

$g\left(x\right)\in \left\{{\mathrm{coth}}^{n}\left(x\right),{\text{sech}}^{n}\left(x\right),\text{csc}{h}^{n}\left(x\right)\right\}$. $\forall n,m\in N$.

3) $\int {\mathrm{coth}}^{m}\left(x\right){\text{sech}}^{n}\left(x\right)\text{d}x$.

Cite this paper: Ali, A. and Abd-Alkanee, O. (2019) The Power Integrations of Trigonometric and Hyperbolic Functions. Open Access Library Journal, 6, 1-15. doi: 10.4236/oalib.1105618.
References

   Thomas Jr., G.B., Weir, M.D. and Hass, J. (2010) Thomas’ Calculus. 12th Edition, Pearson, Boston, New York.

   Wilson, F.C. and Adamson, S. (2009) Applied Calculus. Hough-ton Mifflin Harcourt Publishing Company, Boston, New York.

   Wrede, R. and Spiegel, M.R. (2010) Advance Calculus. 3rd Edition, Schaum’s Outline Series, New York.

   Anton, H., Bivens, L. and Davis, S. (2008) Calculus. 7th Edition, John Wiley and Sons, New York.

   Keisler, H.J. (2010) Elementary Calculus. 2nd Edi-tion.

   Spiegel, M.R., Lipschutz, S. and Liu, J. (2009) Mathematical Handbook of For-mulas and Tables. 3rd Edition, Schaum’s Outline Series, New York.

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