Integration of Trigonometric Functions - Formulas, Solved Examples

Integration of Trigonometric Functions

Inverse trigonometric functions are known as arcus functions, cyclometric functions, or anti-trigonometric functions. These functions are used to get an angle for a given trigonometric value. It refers to the change in the value of the trigonometric function at a certain rate.

Inverse Trigonometric Functions

1. $\frac{d}{d x}(-\cos x)=\sin x \Rightarrow \int \sin x d x=-\cos x+C$
2. $\frac{d}{d x}(\sin x)=\cos x \Rightarrow \int \cos x d x=\sin x+C$
3. $\frac{d}{d x}(\tan x)=\sec ^2 x \Rightarrow \int \sec ^2 x d x=\tan x+C$
4. $\frac{d}{d x}(-\cot x)=\csc ^2 x \Rightarrow \int \csc ^2 x d x=-\cot \mathrm{x}+C$
5. $\frac{\mathrm{d}}{\mathrm{dx}}(\sec \mathrm{x})=\sec x \tan \mathrm{x} \Rightarrow \int \sec x \tan x d x=\sec x+C$
6. $\frac{d}{d x}(-\csc x)=\csc x \cot x \Rightarrow \int \csc x \cot x d x=-\csc x+C$

Integrals of $\tan x, \cot x, \sec x, \operatorname{cosec} x$
7. $\frac{d}{d x}(\log |\sin \mathrm{x}|)=\cot \mathrm{x} \Rightarrow \int \cot \mathrm{xdx}=\log |\sin \mathrm{x}|+\mathrm{C}$
8. $\frac{d}{d x}(-\log |\cos x|)=\tan x \Rightarrow \int \tan x d x=-\log |\cos x|+C$
9. $\frac{\mathrm{d}}{\mathrm{dx}}(\log |\sec \mathrm{x}+\tan \mathrm{x}|)=\sec \mathrm{x} \Rightarrow \int \sec \mathrm{xdx}=\log |\sec \mathrm{x}+\tan \mathrm{x}|+\mathrm{C}$
10. $\frac{\mathrm{d}}{\mathrm{dx}}(\log |\csc \mathrm{x}-\cot x|)=\csc x \Rightarrow \int \csc x d x=\log |\csc x-\cot x|+C$

11. $\begin{aligned}
& \int \sec ^2 x d x=\tan x+C \\
& \int\csc ^2 x d x=-\cot x+C
\end{aligned}$

Recommended Video Based on Integration of Trigonometric Functions

Solved Examples Based On Integration of Trigonometric Functions

Example 1: $\int \frac{\sin ^8 x-\cos ^8 x}{\left(1-2 \sin ^2 x \cos ^2 x\right)} d x$ is equal to I
1) $\frac{1}{2} \sin 2 x+c$
2) $-\frac{1}{2} \sin 2 x+c$
3) $-\frac{1}{2} \sin x+c$
4) $-\sin ^2 x+c$

Solution

As learned in the concept

Integrals for Trigonometric functions -

$\begin{aligned}
& \frac{\mathrm{d}}{\mathrm{d} x}(-\cos x)=\sin x \\
& \therefore \int \sin x d x=-\cos x+c \\
& \int \frac{\sin ^8 x-\cos ^8 x}{\left(1-2 \sin ^2 x \cos ^2 x\right)} d x \\
& \int \frac{\left(\sin ^4-\cos ^4 x\right)\left(\sin ^4 x+\cos ^4 x\right)}{1-2 \sin ^2 x \cos ^2 x} d x \\
& \int \frac{\left(\sin ^2+\cos ^2 x\right)\left(\sin ^2-\cos ^2 x\right)\left(\sin ^4 x+\cos ^4 x\right)}{1-2 \sin ^2 x \cos ^2 x} d x \\
& \int \frac{1\left(\sin ^2-\cos ^2 x\right)\left(\sin ^4 x+\cos ^4 x\right)}{1-2 \sin ^2 x \cos ^2 x} d x \\
& \text { and }\left(\sin ^4 x+\cos ^4 x\right)=\left(\sin ^2 x+\cos ^2 x\right)^2-2\left(\sin ^4 x \cdot \cos ^4 x\right)=1-2\left(\sin ^4 x \cdot \cos ^4 x\right) \\
& \int \frac{1\left(\sin ^2-\cos ^2 x\right)\left(1-2 \sin ^2 x \cos ^2 x\right)}{1-2 \sin ^2 x \cos ^2 x} d x \\
& \int\left(\sin ^2-\cos ^2 x\right) d x \\
& \int-\cos 2 x d x \\
& =\frac{-1}{2} \sin 2 x+C
\end{aligned}$
Hence, the answer is the option 2.

Example 2: The integral $\int \sqrt{1+2 \cot x(\operatorname{cosec} x+\cot x) d x}\left(0<x<\frac{\pi}{2}\right)$ is equal to (where C is a constant of integration)
1)$4 \log \left(\sin \frac{x}{2}\right)+C$

2)$2 \log \left(\sin \frac{x}{2}\right)+C$

3)$2 \log \left(\cos \frac{X}{2}\right)+C$

4)$4 \log \left(\cos \frac{x}{2}\right)+C$

Solution

Integrals for Trigonometric functions -

Here we use trigonometric substitution to change this equation into a fundamental equation of integration.

$\begin{aligned}
& \frac{\mathrm{d}}{\mathrm{d} x}(-\cos x)=\sin x \\
& \therefore \int \sin x d x=-\cos x+c \\
& \int \sqrt{\left(1+2 \cot x \operatorname{cosec} x+2 \cot ^2 x\right) d x} \\
& =\int \sqrt{\left(1+\cot ^2 x\right)+2 \cot x \operatorname{cosec} x+\cot ^2 x} d x \\
& =\int \sqrt{\left(\operatorname{cosec}^2 x+2 \cot x \operatorname{cosec} x+\cot ^2 x\right)} d x \\
& =\int(\operatorname{cosec} x+\cot x) d x \\
& =\ln |\csc x-\cot x|+\ln |\sin x|+c \\
& =\ln |1-\cos x|+c \\
& =\ln \left|2 \sin \frac{x}{2}\right|+c \\
& =\ln \left|\sin ^2 \frac{x}{2}\right|+\ln 2+c \\
& =2 \ln \left|\sin \frac{x}{2}\right|+c_1
\end{aligned}$

Hence, the answer is the option (2).

Example 3: Let $\alpha \epsilon(0, \pi / 2)$ be fixed. If the integral $\int \frac{\tan x+\tan \alpha}{\tan x-\tan \alpha} d x=A(x) \cos 2 \alpha+B(x) \sin 2 \alpha+C$, where C is a constant of integration, then the functions $\mathrm{A}(\mathrm{x})$ and $\mathrm{B}(\mathrm{x})$ are respectively :
1) $x+\alpha$ and $\log _e|\sin (x+\alpha)|$
2) $x-\alpha$ and $\log _e|\sin (x-\alpha)|$
3) $x-\alpha$ and $\log _e|\cos (x-\alpha)|$
4) $x+\alpha$ and $\log _e|\sin (x-\alpha)|$

Solution
Integral of Trigonometric functions -

$\int \cot x d x=\ln |\sin x|+C$

Here,

$\begin{aligned}
& \mathrm{I}=\int \frac{\tan x+\tan \alpha}{\tan x-\tan \alpha} d x=A(x) \cos 2 \alpha+B(x) \sin 2 \alpha+C \\
& I=\int \frac{\frac{\sin x}{\cos x}+\frac{\sin \alpha}{\cos \alpha}}{\frac{\sin x}{\cos x}-\frac{\sin \alpha}{\cos \alpha}} d x \\
& =\int \frac{\sin x \cos \alpha+\sin \alpha \cos x}{\sin x \cos \alpha-\sin \alpha \cos x} d x \\
& =\int \frac{\sin (x+\alpha)}{\sin (x-\alpha)} d x \\
& =\int \frac{\sin (x-\alpha+2 \alpha)}{\sin (x-\alpha)} d x \\
& \text { put } x-\alpha=t ; d x=d t \\
& =\int \frac{\sin (t+2 \alpha)}{\sin (t)} d t \\
& =\cos (2 \alpha) \int d t+\sin (2 \alpha) \int \cot (t) d t \\
& =(x-\alpha) \cos 2 \alpha+\log |\sin (x-\alpha)| \sin 2 \alpha+C
\end{aligned}$

comparing with LHS $\Rightarrow$

$\left[\begin{array}{c}
A(x)=x-\alpha \\
B(x)=\log _e|\sin (x-\alpha)|
\end{array}\right]$

Hence, the answer is the option 2.

Example 4: For $x^2 \neq n \pi+1, n \in N_{\text {(the set of natural numbers), the integral }} \int x \sqrt{\frac{2 \sin \left(x^2-1\right)-\sin 2\left(x^2-1\right)}{2 \sin \left(x^2-1\right)+\sin 2\left(x^2-1\right)}} d x$ is equal to: (where c is a constant of integration) $\log _e\left|\frac{1}{2} \sec \left(x^2-1\right)\right|+c$
2) $\frac{1}{2} \log _e\left|\sec ^2\left(x^2-1\right)\right|+c$
3) $\frac{1}{2} \log _c\left|\sec \left(\frac{x^2-1}{2}\right)\right|+c$
4) $\log _x\left|\sec ^2\left(\frac{x^2-1}{2}\right)\right|+c$

Solution

Integral of Trigonometric functions -

$\int \tan x d x=\ln |\sec x|+C$

$\begin{aligned} & \int \cot x d x=\ln |\sin x|+C \\ & \int \sec x d x=\ln |\sec x+\tan x|+C \\ & \int \operatorname{cosec} x d x=\ln |\operatorname{cosec} x-\cot x|+C \\ & \text { Put }\left(x^2-1\right)=t ; \quad 2 x d x=d t \\ & \int x \sqrt{\frac{2 \sin \left(x^2-1\right)-\sin 2\left(x^2-1\right)}{2 \sin \left(x^2-1\right)+\sin 2\left(x^2-1\right)}} d x=\frac{1}{2} \int \sqrt{\frac{2 \sin (t)-\sin 2 t}{2 \sin (t)+\sin (2 t)}} d t \\ & \because \sin (2 t)=2 \sin (t) \cos (t) \\ & \Rightarrow \frac{1}{2} \int \sqrt{\frac{1-\cos (t)}{1+\cos (t)}} d t \\ & \Rightarrow \frac{1}{2} \int \tan \left(\frac{t}{2}\right) d t \\ & \Rightarrow \ln \left|\sec \left(\frac{t}{2}\right)\right|+C \\ & \text { replace t with }\left(x^2-1\right) \\ & \Rightarrow \ln \left|\sec \left(\frac{x^2-1}{2}\right)\right|+C\end{aligned}$

Hence, the answer is the option 4.

Example 5: If $\int_0^{\frac{\pi}{2}} \frac{\cot x}{\cot x+\operatorname{cosec} x} d x=m(\pi+n)$, then $m \cdot n$ is equal to :

1) -1

2) 1

3) -0.5

4) 0.5

Solution

Integration of trigonometric function -

$\begin{aligned}
& \int \sec ^2 x d x=\tan x+C \\
& \int \csc ^2 x d x=-\cot x+C
\end{aligned}$

$\begin{aligned}
\int_0^{\frac{\pi}{2}} \frac{\cot x}{\cot x+\csc x} d x & =\int_0^{\frac{\pi}{2}} \frac{\frac{\cos x}{\sin x}}{\frac{\cos x}{\sin x}+\frac{1}{\sin x}} d x \\
& =\int_0^{\frac{\pi}{2}} \frac{\cos x}{\cos x+1} d x \\
& =\int_0^{\frac{\pi}{2}} \frac{2 \cos ^2 \frac{x}{2}-1}{2 \cos ^2 \frac{x}{2}-x+x} d x \\
& =\int_0^{\frac{\pi}{2}}\left(1-\frac{1}{2} \sec ^2 \frac{x}{2}\right) d x \\
& =\left[x-\tan \frac{x}{2}\right]_0^{\frac{\pi}{2}} \\
& =\frac{1}{2}[\pi-2]
\end{aligned}$
So, $m=\frac{1}{2} \quad$ and $\quad n=-2$

$\therefore m n=\frac{1}{2} \times-2=-1$

Hence, the answer is the option (1).