Integration of Trigonometric Functions - Formulas, Solved Examples

The integration of trigonometric functions is one of the important parts of calculus, and it is applied to measure the change in the function at a certain point. Mathematically, it forms a powerful tool by which slopes of functions are determined, the maximum and minimum of functions found, and problems on motion, growth, and decay, to name a few. These concepts of integration have been broadly applied in branches of mathematics, physics, engineering, economics, and biology.

In this article, we will cover the concept of Integration. This concept falls under the broader category of Calculus, which is a crucial Chapter in class 12 Mathematics. It is not only essential for board exams but also for competitive exams like the Joint Entrance Examination (JEE Main), and other entrance exams such as SRMJEE, BITSAT, WBJEE, BCECE, and more. Over the last ten years of the JEE Main exam (from 2013 to 2023), a total of seventeen questions have been asked on this concept, including one in 2014, one in 2018, six in 2019, four in 2020, three in 2021, and two in 2023.

What is Integration?

Integration is the reverse process of differentiation. In integration, we find the function whose differential coefficient is given. The rate of change of a quantity y concerning another quantity x is called the derivative or differential coefficient of y concerning x. Geometrically, the Differentiation of a function at a point represents the slope of the tangent to the graph of the function at that point.

Integral is based on a limiting procedure which approximates the area of a curvilinear region by breaking the region into thin vertical slabs.”

For example,

$\begin{aligned} & \frac{d}{d x}(\sin x)=\cos x \\ & \frac{d}{d x}\left(x^2\right)=2 x \\ & \frac{d}{d x}\left(e^x\right)=e^x\end{aligned}$

In the above example, the function cos(x) is the derivative of sin(x). We say that sin(x) is an anti-derivative (or an integral) of cos(x). Similarly, x² and e^x are the antiderivatives (or integrals) of 2x and e^x respectively.

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

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).