The Power Integrations of Trigonometric and Hyperbolic Functions

Ahmed M. Ali^*, Omar M. Abd-Alkanee

The Surveying Department, Department of Mathematics, University of Mosul, Mosul, Iraq.

DOI: 10.4236/oalib.1105618                

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.

Keywords

Trigonometric Functions, Hyperbolic Functions, Integrations Method

1. Introduction

We follow the terminology of [1] [2] [3] . 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 [4] [5] .

Theorem 1.1 [6] : For all $k\in N$. then:

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

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

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

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

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

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

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

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

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

Proof:

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

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

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

For $k=2$. then

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

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

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

For hypotheses induction, we get:

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

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:

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

For $k=2$. then

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

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

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

For hypotheses induction, we get:

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

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:

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

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

Proof:

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

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

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

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

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

Proof: Directed from Theorem (1.1). #

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

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

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

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

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

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

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

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

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

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

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

Using integration by parts:

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

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

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

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

Hence

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

Note: From clearly that:

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

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

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

Proof: Since

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

(by Lemma 2.3). Then

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

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

…

and so on

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

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

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

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

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

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

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

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

Using integration by parts:

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

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

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

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

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

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

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

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

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

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

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

...

and so on

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

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

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

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

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

3. The Integration of Multiply Trigonometric Function Power

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

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

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

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

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

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

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

where c is integral constant.

Proof: Directed proof. #

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

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

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

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

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

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

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

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

By Theorem 2.1, we have (3.3).

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

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

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

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

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

Theorem 3.5: For all $n,m\in N$.nd m is an even number, then: ${\displaystyle \int {\tan }^m(x)}{\sec }^n(x)dx=$

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

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

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

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

By Theorem 2.2, we have (3.5). #

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

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

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

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

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

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

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

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

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

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

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

4. The Hyperbolic Functions

The combinations of the exponential functions $\exp \left(x\right)$.nd $\exp \left(-x\right)$.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) ${\sinh }^{2k-1}\left(x\right)=\frac{1}{2^{2k-2}}{\displaystyle \underset{n=0}{\overset{k-1}{\sum }}{\left(-1\right)}^n\left(\begin{array}{c}2k-1\\ n\end{array}\right)\sinh \left(2k-1-2n\right)x}$.

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

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

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

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

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

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

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

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

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

4.2. The Integration of Hyperbolic Functions Power

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

4.3. The Integration of Multiply Trigonometric Function Power

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

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

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

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

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

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

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

where c is integral constant.

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

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

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

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

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

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

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

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

Remark

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

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

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

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

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

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

Conflicts of Interest

The authors declare no conflicts of interest regarding the publication of this paper.

References

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