Integration of Trigonometric Functions - Formulas, Solved Examples

What is Integration?

Integration is the reverse process of differentiation. In simple terms, it helps us find the original function when its derivative (or rate of change) is known. In differentiation, we determine how a quantity changes; in integration, we reconstruct the quantity from its rate of change.

Geometrically, differentiation represents the slope of the tangent to a curve at a point, while integration represents the area under a curve. The process of integration is based on a limiting procedure, which approximates the area of a curved region by dividing it into infinitesimally small vertical strips and summing their areas.

For example:

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

In the above examples, $\cos x$ is the derivative of $\sin x$. Hence, $\sin x$ is called the antiderivative or integral of $\cos x$. Similarly, $x^2$ and $e^x$ are the antiderivatives (or integrals) of $2x$ and $e^x$ respectively.

Integration of Trigonometric Functions

The integration of trigonometric functions involves finding the antiderivatives of trigonometric ratios such as $\sin x$, $\cos x$, $\tan x$, $\cot x$, $\sec x$, and $\csc x$. These integrals frequently appear in problems involving periodic motion, wave equations, and geometry.

Trigonometric integrals form a fundamental part of Class 12 Mathematics (Chapter 7 – Integrals) and are essential for mastering JEE, CUET, and CBSE board exams.

Standard Integrals of Basic Trigonometric Functions

Here are the most common integration formulas for trigonometric functions derived from their corresponding derivatives:

1. $\frac{d}{dx}(-\cos x) = \sin x \Rightarrow \int \sin x , dx = -\cos x + C$

2. $\frac{d}{dx}(\sin x) = \cos x \Rightarrow \int \cos x , dx = \sin x + C$

3. $\frac{d}{dx}(\tan x) = \sec^2 x \Rightarrow \int \sec^2 x , dx = \tan x + C$

4. $\frac{d}{dx}(-\cot x) = \csc^2 x \Rightarrow \int \csc^2 x , dx = -\cot x + C$

5. $\frac{d}{dx}(\sec x) = \sec x \tan x \Rightarrow \int \sec x \tan x , dx = \sec x + C$

6. $\frac{d}{dx}(-\csc x) = \csc x \cot x \Rightarrow \int \csc x \cot x , dx = -\csc x + C$

Integrals of $\tan x$, $\cot x$, $\sec x$, and $\csc x$

Using logarithmic properties and trigonometric derivatives, we obtain the following standard results:

7. $\frac{d}{dx}(\log |\sin x|) = \cot x \Rightarrow \int \cot x , dx = \log |\sin x| + C$

8. $\frac{d}{dx}(-\log |\cos x|) = \tan x \Rightarrow \int \tan x , dx = -\log |\cos x| + C$

9. $\frac{d}{dx}(\log |\sec x + \tan x|) = \sec x \Rightarrow \int \sec x , dx = \log |\sec x + \tan x| + C$

10. $\frac{d}{dx}(\log |\csc x - \cot x|) = \csc x \Rightarrow \int \csc x , dx = \log |\csc x - \cot x| + C$

Additional Standard Integrals

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

Integration Techniques for Trigonometric Functions

Integration of trigonometric functions often requires strategic methods to simplify complex expressions. These techniques - substitution, parts, reduction formulas, and partial fractions - form the foundation of advanced calculus and are especially useful for JEE, CUET, and Class 12-level problems.

Integration by Substitution

The substitution method is one of the most commonly used techniques for solving trigonometric integrals. It involves transforming a complex function into a simpler one by introducing a new variable.

If $t = g(x)$, then $dt = g'(x),dx$.
Hence, $\int f(g(x))g'(x),dx = \int f(t),dt$

This technique works especially well when the given integral contains composite trigonometric functions or multiple angles.

Examples:

• $\int \sin 2x,dx$: Let $t = 2x \Rightarrow dt = 2dx$

• $\int \cos 3x,dx$: Let $t = 3x \Rightarrow dt = 3dx$

• $\int \tan 2x,dx$: Substitute $t = 2x$ to simplify and integrate easily.

The substitution method simplifies the process by reducing higher-degree trigonometric terms into their standard forms.

Integration by Parts

The integration by parts formula is useful when two functions are multiplied together - especially when one part can be easily differentiated and the other integrated.

The formula is given by: $\int u,dv = uv - \int v,du$

Here, $u$ and $v$ are functions of $x$.
To decide which function to take as $u$, use the ILATE rule (Inverse, Logarithmic, Algebraic, Trigonometric, Exponential).

This method is particularly effective for integrals like:

• $\int x \sin x,dx$

• $\int \theta \cos \theta,d\theta$

• $\int \tan^{-1}x \sin x,dx$

By carefully choosing $u$ and $dv$, complex trigonometric products can be reduced to simpler, solvable integrals.

Integration Using Reduction Formula

A reduction formula expresses an integral containing a power of a trigonometric function in terms of another integral with a lower power. This recursive approach is highly useful for repeated trigonometric powers.

For instance: $\int \sin^n x,dx$ and $\int \cos^n x,dx$ can be written as reduction forms: $\int \sin^n x,dx = -\frac{1}{n}\sin^{n-1}x \cos x + \frac{n-1}{n}\int \sin^{n-2}x,dx$

These formulas help simplify expressions like $\sin^5 x$, $\cos^4 x$, etc., and are essential in deriving definite integrals involving powers 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
$\begin{aligned}
& \frac{\mathrm{d}}{\mathrm{d} x}(-\cos x)=\sin x \\
& \therefore \int \sin x d x=-\cos x+c \\
& \int \frac{\sin ^8 x-\cos ^8 x}{\left(1-2 \sin ^2 x \cos ^2 x\right)} d x \\
& \int \frac{\left(\sin ^4-\cos ^4 x\right)\left(\sin ^4 x+\cos ^4 x\right)}{1-2 \sin ^2 x \cos ^2 x} d x \\
& \int \frac{\left(\sin ^2+\cos ^2 x\right)\left(\sin ^2-\cos ^2 x\right)\left(\sin ^4 x+\cos ^4 x\right)}{1-2 \sin ^2 x \cos ^2 x} d x \\
& \int \frac{1\left(\sin ^2-\cos ^2 x\right)\left(\sin ^4 x+\cos ^4 x\right)}{1-2 \sin ^2 x \cos ^2 x} d x \\
& \text { and }\left(\sin ^4 x+\cos ^4 x\right)=\left(\sin ^2 x+\cos ^2 x\right)^2-2\left(\sin ^4 x \cdot \cos ^4 x\right)=1-2\left(\sin ^4 x \cdot \cos ^4 x\right) \\
& \int \frac{1\left(\sin ^2-\cos ^2 x\right)\left(1-2 \sin ^2 x \cos ^2 x\right)}{1-2 \sin ^2 x \cos ^2 x} d x \\
& \int\left(\sin ^2-\cos ^2 x\right) d x \\
& \int-\cos 2 x d x \\
& =\frac{-1}{2} \sin 2 x+C
\end{aligned}$
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).

List of Topics Related to Integration of Trigonometric Functions

This section provides a detailed overview of all key subtopics linked to the integration of trigonometric functions. It helps you understand how trigonometric identities and formulas are applied in solving different types of integral problems efficiently.

Application of Integrals

Integral of Particular Functions

Indefinite Integrals

Integration by Parts

Application of Inequality in Definite Integration

NCERT Resources

Find comprehensive NCERT-based study materials, including chapter-wise notes, detailed solutions, and exemplar problems from Class 12 Chapter 7 – Integrals, designed to support conceptual clarity and exam preparation.

NCERT Class 12 Maths Notes for Chapter 7 - Integrals

NCERT Class 12 Maths Solutions for Chapter 7 - Integrals

NCERT Class 12 Maths Exemplar Solutions for Chapter 7 - Integrals

Practice Questions based on Integration of Trigonometric Functions

This section includes topic-wise MCQs and practice exercises to test your grasp on the integration of trigonometric functions and related methods.

Integration Of Trigonometric Functions- Practice Question MCQ

We have provided below the practice questions related to different concepts of integration to improve your understanding: