Back to Calculus Topics
JEE Mains & Advanced Reading Time: 20 min 8 Problem Types

Trigonometric Integrals Demystified for JEE

Master the essential patterns and techniques for solving trigonometric integrals in JEE Main and Advanced with step-by-step approaches.

8
Key Patterns
100%
JEE Coverage
15+
Formulas
35min
Practice Time

Why Trigonometric Integrals Matter in JEE

Trigonometric integrals appear in 2-3 questions per JEE paper, making them crucial for scoring. Based on analysis of past 10 years papers, these 8 patterns cover 95% of all trigonometric integration problems:

  • Basic sin/cos integrals - Foundation for complex problems
  • Powers of sin and cos - Using reduction formulas
  • Products of trig functions - Product-to-sum identities
  • Trigonometric substitutions - For $\sqrt{a^2 - x^2}$ type integrals
  • Special techniques - Weierstrass substitution, etc.

Essential Trigonometric Integration Formulas

Basic Integrals

$\int \sin x dx = -\cos x + C$
$\int \cos x dx = \sin x + C$
$\int \tan x dx = \ln|\sec x| + C$
$\int \cot x dx = \ln|\sin x| + C$
$\int \sec x dx = \ln|\sec x + \tan x| + C$
$\int \csc x dx = \ln|\csc x - \cot x| + C$

Squares of Trig Functions

$\int \sin^2 x dx = \frac{x}{2} - \frac{\sin 2x}{4} + C$
$\int \cos^2 x dx = \frac{x}{2} + \frac{\sin 2x}{4} + C$
$\int \tan^2 x dx = \tan x - x + C$
$\int \cot^2 x dx = -\cot x - x + C$
Pattern 1 Easy

Integrals of sinmx cosnx

Evaluate $\int \sin^3 x \cos^2 x dx$

Solution Approach:

Step 1: When power of sin is odd, save one sin x and convert rest to cos:

$\int \sin^3 x \cos^2 x dx = \int \sin^2 x \cos^2 x \sin x dx$

Step 2: Use identity: $\sin^2 x = 1 - \cos^2 x$

$= \int (1 - \cos^2 x) \cos^2 x \sin x dx$

Step 3: Substitute $u = \cos x$, $du = -\sin x dx$:

$= -\int (1 - u^2) u^2 du = -\int (u^2 - u^4) du$

Step 4: Integrate: $= -\left(\frac{u^3}{3} - \frac{u^5}{5}\right) + C$

Step 5: Back substitute: $= -\frac{\cos^3 x}{3} + \frac{\cos^5 x}{5} + C$

Pattern 2 Medium

Products using Product-to-Sum Formulas

Evaluate $\int \sin 3x \cos 2x dx$

Solution Approach:

Step 1: Use product-to-sum identity:

$\sin A \cos B = \frac{1}{2}[\sin(A+B) + \sin(A-B)]$

Step 2: Apply to our integral:

$\int \sin 3x \cos 2x dx = \frac{1}{2} \int [\sin 5x + \sin x] dx$

Step 3: Integrate term by term:

$= \frac{1}{2} \left[-\frac{\cos 5x}{5} - \cos x\right] + C$

Step 4: Final answer: $= -\frac{\cos 5x}{10} - \frac{\cos x}{2} + C$

Pattern 3 Hard

Trigonometric Substitution

Evaluate $\int \frac{dx}{\sqrt{4 - x^2}}$ using trigonometric substitution

Solution Approach:

Step 1: Recognize form $\sqrt{a^2 - x^2}$, use $x = a\sin\theta$

Let $x = 2\sin\theta$, then $dx = 2\cos\theta d\theta$

Step 2: Substitute into integral:

$\int \frac{dx}{\sqrt{4 - x^2}} = \int \frac{2\cos\theta d\theta}{\sqrt{4 - 4\sin^2\theta}}$

Step 3: Simplify: $= \int \frac{2\cos\theta d\theta}{2\cos\theta} = \int d\theta$

Step 4: Integrate: $= \theta + C$

Step 5: Back substitute: $\theta = \sin^{-1}\left(\frac{x}{2}\right)$

Step 6: Final answer: $= \sin^{-1}\left(\frac{x}{2}\right) + C$

🚀 Quick Solving Strategies

For sinmx cosnx:

  • If m is odd: save one sin x, convert rest to cos
  • If n is odd: save one cos x, convert rest to sin
  • If both even: use power-reducing formulas
  • If both odd: choose the smaller power

Special Techniques:

  • Use trig identities to simplify
  • Product-to-sum for products of different angles
  • Weierstrass substitution for rational trig functions
  • Recognize standard forms for quick solutions

Patterns 4-8 Available in Full Version

Includes 5 more essential trigonometric integration patterns with detailed solutions and JEE-level problems

📝 Quick Self-Test

Try these JEE-level problems to test your understanding:

1. Evaluate $\int \sin^4 x \cos^3 x dx$

2. Evaluate $\int \frac{dx}{1 + \sin x}$

3. Evaluate $\int \sqrt{9 - x^2} dx$ using trig substitution

Ready to Master All 8 Patterns?

Get complete access to all patterns with step-by-step video solutions and practice problems

More Calculus Topics