Mastering ∫ sin^m(x) cos^n(x) dx — A Complete Guide for JEE Main

Introduction

Integrals of the form ∫ sin^m(x) cos^n(x) dx are a staple in JEE Main and Advanced. Appearing both as direct questions and crucial intermediate steps, fluency with these techniques saves time and builds confidence. This guide systematically breaks down the three core strategies, supported by solved examples and previous JEE questions.

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Strategy 1: One Power is Odd — The Substitution Method

Core Idea

When m or n is an odd positive integer, factor out one trigonometric function to pair with dx, then substitute using the Pythagorean identity.

Procedure

• If m (power of sin x) is odd: Let u = cos x, then -sin x dx = du.
  Use sin²x = 1 – cos²x to convert the remaining even power of sin x.
• If n (power of cos x) is odd: Let u = sin x, then cos x dx = du.
  Use cos²x = 1 – sin²x to convert the remaining even power of cos x.

Example 1: ∫ sin³x cos⁴x dx

Here, sin x has an odd power (3). Let u = cos x ⇒ du = –sin x dx.

$\begin{aligned} \int \sin^3 x \cos^4 x \, dx &= \int \sin^2 x \cdot \cos^4 x \cdot (\sin x \, dx) \\ &= \int (1 - \cos^2 x) \cos^4 x \cdot (\sin x \, dx) \\ &= -\int (1 - u^2) u^4 \, du \\ &= -\int (u^4 - u^6) \, du \\ &= -\left( \frac{u^5}{5} - \frac{u^7}{7} \right) + C \\ &= \boxed{-\frac{\cos^5 x}{5} + \frac{\cos^7 x}{7} + C} \end{aligned}$

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Strategy 2: Both Powers are Even — Power-Reduction Formulas

Core Idea

When m and n are both even, repeatedly apply the half-angle identities to reduce the powers until a simple trigonometric integral is obtained.

Essential Identities

$\sin^2 x = \frac{1 - \cos 2x}{2}, \quad \cos^2 x = \frac{1 + \cos 2x}{2}, \quad \sin x \cos x = \frac{\sin 2x}{2}$

Example 2: ∫ sin⁴x cos²x dx

Both powers are even.

$\begin{aligned} \int \sin^4 x \cos^2 x \, dx &= \int (\sin^2 x)^2 \cos^2 x \, dx \\ &= \int \left( \frac{1 - \cos 2x}{2} \right)^2 \left( \frac{1 + \cos 2x}{2} \right) dx \\ &= \frac{1}{8} \int (1 - \cos 2x)^2 (1 + \cos 2x) \, dx \\ &= \frac{1}{8} \int (1 - \cos 2x - \cos^2 2x + \cos^3 2x) \, dx \quad \text{(Expand carefully)} \\ &\text{After simplification and integrating term by term:} \\ &= \boxed{\frac{x}{16} - \frac{\sin 4x}{64} + \frac{\sin^3 2x}{48} + C} \end{aligned}$ Note: The key is patient expansion and applying power-reduction again if needed.

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Strategy 3: (m + n) is a Negative Even Integer — The Tangent Substitution

Core Idea

When m + n is a negative even integer (e.g., –2, –4), the integrand can often be expressed in terms of tan x and sec²x. The substitution tan x = t then rationalizes the integral.

Why It Works

$\tan x = t \quad \Rightarrow \quad \sec^2 x \, dx = dt \quad \text{and} \quad \sec^2 x = 1 + t^2$ Expressing sin x and cos x in terms of t converts the integral into a rational function in t.

Example 3: ∫ dx / (sin⁴x cos²x)

Here, m = –4, n = –2 ⇒ m + n = –6 (negative even integer).

\begin{aligned} \int \frac{dx}{\sin^4 x \cos^2 x} &= \int \frac{\sec^6 x}{\tan^4 x} \, dx \quad \text{(Multiply numerator & denominator by\sec^6 x)} \\ &= \int \frac{(1 + \tan^2 x)^2 \sec^2 x}{\tan^4 x} \, dx \end{aligned} Let t = tan x ⇒ dt = sec²x dx. $= \int \frac{(1 + t^2)^2}{t^4} \, dt = \int (t^{-4} + 2t^{-2} + 1) \, dt$ $= -\frac{1}{3t^3} - \frac{2}{t} + t + C$ $= \boxed{-\frac{1}{3\tan^3 x} - 2\cot x + \tan x + C}$

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JEE Main Previous Year Questions (PYQs) – Solved

PYQ 1 (JEE Main 2021): A Definite Integral Trick

Evaluate: $\displaystyle I = \int_0^{\pi/2} \frac{\sin^3 x}{\sin^3 x + \cos^3 x} \, dx$

Solution: Use the property $\int_0^a f(x)\,dx = \int_0^a f(a-x)\,dx$.

$I = \int_0^{\pi/2} \frac{\sin^3\left(\frac{\pi}{2} - x\right)}{\sin^3\left(\frac{\pi}{2} - x\right) + \cos^3\left(\frac{\pi}{2} - x\right)} \, dx = \int_0^{\pi/2} \frac{\cos^3 x}{\cos^3 x + \sin^3 x} \, dx$

Adding the original and transformed integrals: $2I = \int_0^{\pi/2} \frac{\sin^3 x + \cos^3 x}{\sin^3 x + \cos^3 x} \, dx = \int_0^{\pi/2} 1 \, dx = \frac{\pi}{2}$ $\boxed{I = \frac{\pi}{4}}$

PYQ 2: Combining Strategies

Evaluate: $\displaystyle \int \frac{\sin^5 x}{\cos^4 x} \, dx$

Solution: Here, sin x has an odd power (5). Let u = cos x ⇒ du = –sin x dx. $\begin{aligned} \int \frac{\sin^5 x}{\cos^4 x} \, dx &= \int \frac{(\sin^2 x)^2 \cdot \sin x}{\cos^4 x} \, dx \\ &= \int \frac{(1 - \cos^2 x)^2}{\cos^4 x} \cdot (\sin x \, dx) \\ &= -\int \frac{(1 - u^2)^2}{u^4} \, du \\ &= -\int (u^{-4} - 2u^{-2} + 1) \, du \\ &= \frac{1}{3u^3} - \frac{2}{u} - u + C \\ &= \boxed{\frac{1}{3\cos^3 x} - 2\sec x - \cos x + C} \end{aligned}$

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Essential Formulas & Identities

Power-Reduction & Simplification

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Common Pitfalls & How to Avoid Them

1. Incorrect Substitution: If sin x has an odd power, substitute for cos x. Don't choose the function whose power is odd.
2. Sign Errors: When setting $u = \cos x$, remember $du = -\sin x \, dx$. It's easy to drop the negative.
3. Overlooking the Hidden Trick: For integrals like $\int \frac{dx}{\sin^4 x \cos^2 x}$, always check the sum of the exponents $m+n$. If it's a negative even integer, $t = \tan x$ is the fastest route.
4. Incomplete Reduction: When both powers are even, apply the half-angle formulas repeatedly until all powers are linear in $\cos(kx)$.
5. Forgetting $+C$: Always add the constant of integration in indefinite integrals.

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Practice Problems for Self-Assessment

1. $\displaystyle \int \sin^3 x \cos^5 x \, dx$
2. $\displaystyle \int \sin^2 x \cos^4 x \, dx$
3. $\displaystyle \int \frac{dx}{\sin^2 x \cos^4 x}$ (Hint: m+n = –6)
4. $\displaystyle \int_0^{\pi/2} \sin^4 x \, dx$
5. $\displaystyle \int \frac{\sin^3 x}{\sqrt{\cos x}} \, dx$

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Final Strategic Summary

To solve $\int \sin^m(x) \cos^n(x) \, dx$ efficiently:

1. Step 1: Check the exponents.
2. Step 2:
   • If m or n is odd: Use substitution. Let $u$ be the trigonometric function with the even power.
   • If both m and n are even: Use power-reduction/half-angle formulas.
   • If m+n is a negative even integer: Use the substitution $t = \tan x$.
3. Step 3: For definite integrals over $[0, \pi/2]$, remember the property $\int_0^{a} f(x)dx = \int_0^{a} f(a-x)dx$ to simplify tricky integrands.

Mastery of these patterns, combined with practice on past JEE problems, will make these integrals a source of marks, not stress.

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Best wishes for your JEE preparation! 🚀