In mathematics, the branch trigonometry is a study all about triangles. Trigonometric integrals are a family of integrals in which trigonometric functions are involved. The three most commonly used functions in trigonometry are Sine, Cosine and Tangent and their respective inverse functions are arcsine, arccosine and arctangent. This page will help you in understanding integral trigonometric functions and inverse trigonometric integrals.


There are number of the basic trigonometric integrals. For example for each trigonometric differential rule, there is a corresponding integration rule,

$\frac{d}{dx}$
sin t = cos t $\frac{dt}{dx}$ is the integration rule $\int$ cos t dt = sin t + C.

Given below is a list of integration rule for the integral trigonometric functions.

$\int (ax + b)^{n}$dx = $\frac{(ax + b)^{n+1}}{a(n+1)}$ + c n $\neq$ -1

$\int$$\frac{1}{\sqrt{1 - x^{2}}}$dx = sin$^{-1}$x + c

$\int$$\frac{f'(x)}{\sqrt{a^{2} - f(x)^{2}}}$dx = $sin^{-1}$[$\frac{f(x)}{a}$] + c

$\int_{a}^{b}$f(x)dx = $\int_{a}^{b}$f(a+b -x)dx

$\int_{a}^{b}f(x)$dx = $\int_{a}^{b}$f(t)dt

Given below is an example for integral trigonometric functions.

Example 1 : Prove that $\int_{0}^{a}$$\frac{1}{x + \sqrt{a^{2} - x^{2}}}$dx = $\frac{\pi}{4}$

Solution :
I = $\int_{0}^{a}$$\frac{1}{x+ \sqrt{a^{2} - x^{2}}}$dx

put x =  a sin $\theta$

$\frac{dx}{d\theta}$ = a cos$\theta$

dx = a cos$\theta d\theta$

when x = 0  => $\theta$ = sin$^{-1}$$\frac{x}{a}$ = sin$^{-1}$0 = 0

when x = a  => $\theta$ = sin$^{-1}$$\frac{a}{a}$ = sin$^{-1}$1 = $\frac{\pi}{2}$

I = $\int_{0}^{\frac{\pi}{2}}$$\frac{acos\theta d\theta}{asin\theta + \sqrt{a^{2}-a^{2}sin^{2}\theta}}$

= $\int_{0}^{\frac{\pi}{2}}\frac{acos\theta d\theta}{asin\theta + \sqrt{a^{2}(1-sin^{2}\theta)}}$

= $\int_{0}^{\frac{\pi}{2}}\frac{acos\theta d\theta}{asin\theta + acos\theta}$

= $\int_{0}$$^{\frac{\pi}{2}}$$\frac{acos\theta d\theta}{a(sin\theta + cos\theta}$

= $\int_{0}^{\frac{\pi}{2}}\frac{cos\theta d\theta}{sin\theta + cos\theta}$    ...................... (1)

I= $\int_{0}$$^{\frac{\pi}{2}}\frac{cos(\frac{\pi}{2}- \theta)}{sin(\frac{\pi}{2}-\theta)+ cos(\frac{\pi}{2}- \theta)}$

I = $\int_{0}^{\frac{\pi}{2}}$$\frac{sin \theta d \theta} {cos \theta + sin \theta}$  ....... (2)

Adding equations, (1) and (2) gives

 I + I = $\int_{0}^{\frac{\pi}{2}}$$\frac{cos\theta d\theta}{cos\theta + sin \theta}$ +  $\int_{0}^{\frac{\pi}{2}}$ $\frac{sin\theta d\theta}{cos\theta + sin \theta}$

2I = $\int_{0}^{\frac{\pi}{2}}$ $\frac{cos\theta + sin \theta }{cos\theta + sin \theta}$ d$\theta$

2I = $\int_{0}^{\frac{\pi}{2}}1 d\theta$

2I = $\theta]_{0}^{\frac{\pi}{2}}$

2I = $\frac{\pi}{2}$ - 0

2I = $\frac{\pi}{2}$

I = $\frac{\pi}{4}$

$\int_{0}^{a}\frac{1}{x+ \sqrt{a^{2} - x^{2}}}$dx = $\frac{\pi}{4}$
Inverse trigonometric functions are the inverse values of trigonometric functions given as sin$^{-1}$, cos$^{-1}$, tan$^{-1}$, sec$^{-1}$, cosec$^{-1}$ and cot$^{-1}$. They are defined as the values of the angles and are used to solve trigonometric equations and its applications.

Below are listed some inverse trigonometric identities:

sin$^{-1}$x + cos$^{-1}$x = $\frac{\pi}{2}$

tan$^{-1}$x + cot$^{-1}$x = $\frac{\pi}{2}$

sec$^{-1}$x + cosec$^{-1}$x = $\frac{\pi}{2}$

tan$^{-1}$x + tan$^{-1}$y = tan$^{-1}$$(\frac{x + y}{1-xy})$

tan$^{-1}$x - tan$^{-1}$y = tan$^{-1}$ $(\frac{x - y}{1+xy})$

2tan$^{-1}$x = tan$^{-1}$($\frac{2x}{1-x^{2}}$) = sin$^{-1}$$(\frac{2x}{1+x^{2}})$ = cos$^{-1}$($\frac{1 - x^{2}}{1 + x^{2}}$)

$\int x^{2}$ arcsin(ax)dx = $\frac{x^{3}arcsin(ax)}{3}$ + $\frac{(a^{2}x^{2}+2)\sqrt{1-a^{2}x^{2}}}{9a^{3}}$ + C

x$^{m}$arccos(ax)dx  = $\frac{x^{m+1}arccos(ax)}{m+1}$ + $\frac{a}{m+1}$$\int$  $\frac{x^{m+1}}{\sqrt{1-a^{2}x^{2}}}$dx   m $\neq$ -1

$\int x^{2}$arctan(ax)dx = $\frac{x^{3}arctan(ax)}{3}$ + $\frac{ln(a^{2}x^{2}+1)}{6a^{3}}$ - $\frac{x^{2}}{6a}$  + C

$\int x^{3}$arccot(ax)dx = $\frac{x^{m+1}arccot(ax)}{m+1}$ + $\frac{a}{m+1}$ $\int$ $\frac{x^{m+1}}{a^{2}x^{2}+1 }$dx (m $\neq$ -1)

$\int x^{3}$arcsec(ax)dx = $\frac{x^{m+1}arcsec(ax)}{m+1}$ - $\frac{1}{a(m+1)}$ $\int$  $\frac{x^{m-1}}{\sqrt{1- \frac{1}{a^{2}x^{2}}}}$dx (m $\neq$ -1)

$\int x^{2}$arccsc(ax)dx = $\frac{x^{3}arccsc(ax)}{3}$ + $\frac{1}{6a^{3}}artanh\sqrt{1-\frac{1}{a^{2}x^{2}}}$ + $\frac{x^{2}}{6a}\sqrt{1-\frac{1}{a^{2}x^{2}}}$ + C

tan$^{-1}$($\frac{1-x}{1+x}$) = $\frac{\pi}{4}$ - tan$^{-1}$x

tan$^{-1}$($\frac{1+x}{1-x}$) = $\frac{\pi}{4}$ + $tan^{-1}$x
Given below are some problems based on integral trigonometric functions and inverse trigonometric integrals:

Example 1 : Evaluate $\int_{0}^{\pi}$$\frac{sin^{4}\theta }{(1+cos\theta )^{2}}$ d$\theta $

Solution :
We know that  1 + cos$\theta$ = 2cos$^{2}$$\frac{\theta }{2}$ and sin $\theta$ =  2sin $\frac{\theta}{2}$ cos $\frac{\theta}{2}$

$\frac{sin^{4}\theta }{(1+cos\theta )^{2}}$ = $\frac{2^{4}sin^{4}\frac{\theta }{2}cos^{4}\frac{\theta }{2}}{2^{2}cos^{4}\frac{\theta }{2}}$

= 4 sin$^{4}$$\frac{\theta }{2}$

=> $\int_{0}^{\pi}$ $\frac{sin^{4}\theta }{(1+cos\theta )^{2}}d\theta $ = 4$\int_{0}^{\pi}sin^{4}$$\frac{\theta }{2}$d$\theta$

Put $\frac{\theta }{2}$ = t

then d$\theta$ = 2dt
Also when $\theta$ = 0,  t = 0 and when t = $\pi$,  t = $\frac{\pi}{2}$

Now 4 $\int_{0}^{\frac{\pi}{2}}sin^{4}$t dt = 4 $\times$ $\frac{3}{4}$ $\times$ $\frac{1}{2}$ $\times$ $\frac{\pi}{2}$

= $\frac{3}{4}$$\pi$

Example 2 : Evaluate $\int_{0}^{1}$$\frac{sin^{-1}x}{x}$dx

Solution : Let x = sin$\theta$ so that dx = cos$\theta  d\theta $

Now x = 0, $\theta$ = 0
x = 1, $\theta$ = $\frac{\pi}{2}$

Hence I = $\int_{0}^{1}$$\frac{sin^{-1}x}{x}$dx = $\int_{0}^{\frac{\pi}{2}}$ $\frac{\theta cos\theta d\theta }{sin \theta }$
= $\int_{0}^{\frac{\pi}{2}}\theta cot\theta d\theta$

Integrating by parts we get

I = $\theta logsin\theta ]_{0}^{\frac{\pi}{2}}- \int_{0}^{\frac{\pi}{2}}log sin \theta d\theta$

= - $\int_{0}^{\frac{\pi}{2}}log sin\theta d\theta$

 = - (- $\frac{\pi}{2}$ log 2)

= ($\frac{\pi}{2}$)log 2