Trigonometric Integrals

Trigonometric Integrals - Concepts, Rules, and Examples

Trigonometric integrals play a key role in calculus, especially when solving problems involving periodic motion, oscillations, or wave patterns. Whether it’s calculating the total distance covered by a pendulum or analyzing alternating current in physics, integrating trigonometric functions helps determine quantities like area, displacement, and average value over time.

In simple terms, integration is the reverse process of differentiation - it finds the original function from its rate of change. Geometrically, while differentiation gives the slope of a tangent to a curve, integration finds the area under the curve. Trigonometric integrals involve functions like $\sin x$, $\cos x$, and $\tan x$, often requiring clever substitutions to simplify the process.

Basic Understanding of Trigonometric Integrals

Before diving into formulas, remember that these functions can often be transformed using trigonometric identities or substitutions such as $\tan x = t$ or $\tan (x/2) = t$.
Such substitutions convert complex trigonometric expressions into algebraic forms that are easier to integrate.

(a) Integrals of the Form $\int \frac{1}{a \cos^2 x + b \sin^2 x} , dx$

These types of integrals frequently occur when dealing with mixed trigonometric denominators. Examples include:

1. $\int \frac{1}{a \cos^2 x + b \sin^2 x} , dx$

2. $\int \frac{1}{a + b \sin^2 x} , dx$

3. $\int \frac{1}{a + b \cos^2 x} , dx$

4. $\int \frac{1}{a + b \sin^2 x + c \cos^2 x} , dx$

Working Rule

Step 1: Divide both numerator and denominator by $\cos^2 x$.
This converts all terms into $\tan x$ and $\sec x$ form.

Step 2: Substitute $\tan x = t$, which gives $\sec^2 x , dx = dt$.

After substitution, the integral transforms into an algebraic form:
$\int \frac{f(t) , dt}{A t^2 + B t + C}$

Here, $f(t)$ is a polynomial in $t$. This algebraic integral can then be evaluated using standard techniques discussed in earlier integration concepts.

(b) Integrals of the Form $\int \frac{1}{a \sin x + b \cos x} , dx$

These integrals involve linear combinations of sine and cosine functions. Examples include:

1. $\int \frac{1}{a \sin x + b \cos x} , dx$

2. $\int \frac{1}{a + b \sin x} , dx$

3. $\int \frac{1}{a + b \cos x} , dx$

4. $\int \frac{1}{a \sin x + b \cos x + c} , dx$

Working Rule

To simplify such expressions, use the Weierstrass Substitution method:

Write $\sin x$ and $\cos x$ in terms of $\tan \frac{x}{2}$:

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

Now substitute $\tan (x/2) = t$, which gives $dx = \frac{2 , dt}{1 + t^2}$.

After substitution, the integral reduces to the algebraic form:
$\int \frac{1}{a t^2 + b t + c} , dt$

This integral can be easily solved using the quadratic denominator methods covered in earlier topics.

(c) Integrals of the Form $\int \frac{p \cos x + q \sin x + r}{a \cos x + b \sin x + c} , dx$

These are among the most general types of trigonometric integrals, where both the numerator and denominator involve linear trigonometric combinations. Examples include:

1. $\int \frac{p \cos x + q \sin x + r}{a \cos x + b \sin x + c} , dx$

2. $\int \frac{p \cos x + q \sin x}{a \cos x + b \sin x} , dx$

Working Rule

To simplify, express the numerator as a linear combination of the denominator and its derivative.

Let $(p \cos x + q \sin x + r) = \lambda (a \cos x + b \sin x + c) + \mu (-a \sin x + b \cos x) + \gamma$

Here, $\lambda$, $\mu$, and $\gamma$ are constants determined by comparing the coefficients of $\sin x$, $\cos x$, and constant terms on both sides.

Now, the integral becomes:

$\begin{aligned} \int \frac{p \cos x + q \sin x + r}{a \cos x + b \sin x + c} \, dx &= \int \frac{\lambda (a \cos x + b \sin x + c) + \mu (-a \sin x + b \cos x) + \gamma}{a \cos x + b \sin x + c} \, dx \\ &= \lambda \int dx + \mu \int \frac{-a \sin x + b \cos x}{a \cos x + b \sin x + c} \, dx + \gamma \int \frac{dx}{a \cos x + b \sin x + c} \end{aligned}$

Now, using the substitution $a \cos x + b \sin x + c = t$, the second integral simplifies, leading to:

$\int \frac{p \cos x + q \sin x + r}{a \cos x + b \sin x + c} , dx = \lambda x + \mu \ln |a \cos x + b \sin x + c| + \int \frac{\gamma , dx}{a \cos x + b \sin x + c}$

Final Simplified Result

Thus, the final integrated form can be written as:

$\int \frac{p \cos x + q \sin x + r}{a \cos x + b \sin x + c} , dx = \lambda x + \mu \ln |a \cos x + b \sin x + c| + C$

where $\lambda$ and $\mu$ are constants determined by comparing coefficients, and $C$ is the constant of integration.

Trigonometric Integrals Formulae

Below are the important trigonometric integrals formulae, which you must understand to solve complex questions:

Solved Questions Based on Trigonometric Integrals

Example 1: Integrate $\int \frac{d x}{\sin ^2 x+2 \sin x \cos x}$

1) $\ln \frac{t+1}{t-1}+c$
2) $\ln \frac{t+2}{t}+c$
3) $\frac{1}{2} \ln \frac{t}{(t+2)}+C$
4) none of these

Solution
Divide by $\cos ^2 x$ in each case and pull $t=\tan x, d t=\sec ^2 x d x$

$\int \frac{d x \sec ^2 x}{\tan ^2 x+2 \tan x}=\int \frac{d t}{(t+1)^2-1}=\frac{1}{2} \ln \frac{t}{(t+2)}+C$

Hence, the answer is the option 3.

Example 2: $\int \frac{d x}{1+\sin x}$
1) $\frac{1}{1+\tan \left(\frac{x}{2}\right)}+C$

2) $-\frac{2}{\tan \left(\frac{x}{2}\right)+1}+C$
3) $\frac{1}{1+\cot \left(\frac{x}{2}\right)}+C$
4) $-\frac{1}{1+\cot \left(\frac{x}{2}\right)}+C$

Solution

$\begin{aligned}
& I=\int \frac{1}{\sin (x)+1} \mathrm{~d} x \\
& =\int \frac{\sec ^2\left(\frac{x}{2}\right)}{\left(\tan \left(\frac{x}{2}\right)+1\right)^2} \mathrm{~d} x \\
& \text { Put } u=\tan \left(\frac{x}{2}\right)+1 \Rightarrow \mathrm{d} x=\frac{2}{\sec ^2\left(\frac{x}{2}\right)} \mathrm{d} u \\
& I=2 \int \frac{1}{u^2} d u \\
& =-\frac{2}{u} \\
& =-\frac{2}{\tan \left(\frac{x}{2}\right)+1}+C
\end{aligned}$

Hence, the answer is the option (2).

Example 3: $\int \frac{d x}{3+4 \cos ^2 x}$ equals
1) $\frac{\sqrt{3}}{\sqrt{7}} \tan ^{-1}\left(\sqrt{\frac{3}{7}} \tan x\right)+C$
2) $\frac{1}{\sqrt{21}} \tan ^{-1}\left(\sqrt{\frac{3}{7}} \tan x\right)+C$

$\sqrt{3} \tan ^{-1}\left(\sqrt{\frac{3}{7}} \tan x\right)+C$

4) None of these

Solution
Divide numerator and denominator by $\cos ^2 x$

$\begin{aligned}
& \int \frac{\sec ^2 x d x}{3 \sec ^2 x+4} \\
& =\int \frac{\sec ^2 x d x}{3 \tan ^2 x+3+4} \\
& =\int \frac{\sec ^2 x d x}{7+3 \tan ^2 x}
\end{aligned}$

Put $\tan x=t \Rightarrow \sec ^2 x d x=d t$

$\begin{aligned}
& \int \frac{d t}{7+3 t^2} \\
& =\frac{1}{3} \int \frac{d t}{t^2+\left(\sqrt{\frac{7}{3}}\right)^2} \\
& =\frac{1}{3} \times \frac{\sqrt{3}}{\sqrt{7}} \tan ^{-1}\left(\sqrt{\frac{3}{7}} \tan x\right)+C \\
& =\frac{1}{\sqrt{21}} \tan ^{-1}\left(\sqrt{\frac{3}{7}} \tan x\right)+C
\end{aligned}$

Hence, the answer is the option 2.

Example 4: Integrate $\int \frac{d x}{\sin ^2 x+2 \cos ^2 x+2 \sin x \cos x}$
1) $x+c$
2) $\frac{1}{2} \tan ^{-1}(1+x)+C$
3) $\tan ^{-1}(1+\tan x)+C$

4) None of these

Solution

Divide by $\cos^{2} x$

$\begin{aligned}
& \int \frac{\sec ^2 x \cdot d x}{\tan ^2 x+2+2 \tan x} \\
& I=\int \frac{\sec ^2 x d x}{(\tan x+1)^2+1}
\end{aligned}$

Put $\tan (x)=t$

$\sec ^2 x d x=d t$

$\begin{aligned}
& I=\int \frac{d t}{(t+1)^2+1} \\
& =\tan ^{-1}(t+1)+C \\
& =\tan ^{-1}(\tan x+1)+C
\end{aligned}$

Hence, the answer is the option 3.

Example 5: $\int \frac{d x}{1+2 \cos x}$
1) $\frac{1}{\sqrt{3}} \ln \left|\frac{\sqrt{3}+\tan \frac{x}{2}}{\sqrt{3}-\tan \frac{x}{2}}\right|+c$
2) $-\frac{1}{\sqrt{3}} \cdot \ln \left|\frac{\tan \left(\frac{x}{2}\right)-\sqrt{3}}{\tan \left(\frac{x}{2}\right)+\sqrt{3}}\right|+c$
3) $\frac{2}{\sqrt{3}} \ln \left|\frac{\sqrt{3}+\tan \frac{x}{2}}{\sqrt{3}-\tan \frac{x}{2}}\right|+c$
4) None of these

Solution

$\begin{aligned} & \int \frac{d x}{1+2 \cos x} \\ & \int \frac{d x}{1+\frac{2-2 \tan ^2 \frac{x}{2}}{1+\tan ^2 \frac{2}{2}}} \\ & =\int \frac{\sec ^2\left(\frac{x}{2}\right) \cdot d x}{3-\tan ^2\left(\frac{x}{2}\right)} \\ & \text { Put } u=\tan \left(\frac{x}{2}\right) \Rightarrow \frac{1}{2} \sec ^2\left(\frac{x}{2}\right) \cdot d x=d u \\ & =-2 \int \frac{1}{u^2-3} d u \\ & =-\frac{1}{\sqrt{3}} \cdot \ln \left|\frac{u-\sqrt{3}}{u+\sqrt{3}}\right|+c \\ & =-\frac{1}{\sqrt{3}} \cdot \ln \left|\frac{\tan \left(\frac{x}{2}\right)-\sqrt{3}}{\tan \left(\frac{x}{2}\right)+\sqrt{3}}\right|+c\end{aligned}$Hence, the answer is the option (2).

List of Topics Related to Trigonometric Integrals

This section outlines all the key concepts and formulas associated with trigonometric integrals, including topics like integration of particular functions, indefinite integrals, integration by parts, etc.

Application of Integrals

Integral of Particular Functions

Indefinite Integrals

Integration by Parts

Application of Inequality in Definite Integration

NCERT Resources

Here you’ll find direct links and references to NCERT-based materials-notes, solutions, and exemplars-to strengthen your conceptual understanding of integrals.

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 trigonometric integrals

This section provides a wide range of practice questions and MCQs focused on integration involving trigonometric functions. It helps you strengthen concepts like $\sin x$, $\cos x$, $\tan x$, integrals, trigonometric substitutions, and standard reduction formulas that frequently appear in board and competitive exams.

Trigonometric Integrals- Practice Question MCQ

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