The Hook: The Oscillating Signal
In Oppenheimer (2023), scientists study wave patterns—sound waves, light waves, even quantum wave functions. All these waves are described by sine and cosine functions.
Now imagine you have a complex wave pattern like:
$$y = \sin^3 x \cos^2 x$$Question: How do you find the total area under one complete wave cycle?
This requires trigonometric integration—one of the trickiest but most elegant topics in calculus. In JEE, this appears in 2-3 questions every year, often combined with definite integrals.
Why it matters: Trig integration is the foundation for Fourier analysis (used in signal processing, audio engineering, and even MP3 compression)!
Essential Trigonometric Identities
Before we dive into integration, let’s refresh the key identities you’ll need.
Power-Reduction Formulas
Memory Trick: “Sin Subtracts, Cos Adds” (in the numerators)
Product-to-Sum Formulas
$$\sin A \cos B = \frac{1}{2}[\sin(A+B) + \sin(A-B)]$$ $$\cos A \sin B = \frac{1}{2}[\sin(A+B) - \sin(A-B)]$$ $$\cos A \cos B = \frac{1}{2}[\cos(A+B) + \cos(A-B)]$$ $$\sin A \sin B = \frac{1}{2}[\cos(A-B) - \cos(A+B)]$$Triple Angle Formulas
$$\sin^3 x = \frac{3\sin x - \sin 3x}{4}$$ $$\cos^3 x = \frac{3\cos x + \cos 3x}{4}$$Strategy Guide: Choosing the Right Technique
Type 1: $\int \sin^m x \cos^n x \, dx$
- If $m$ or $n$ is odd: Use substitution (save one factor, convert the rest)
- If both $m$ and $n$ are even: Use power-reduction formulas
Type 2: $\int \tan^m x \sec^n x \, dx$
- If $n$ is even: Peel off $\sec^2 x$, use $\tan^2 x = \sec^2 x - 1$
- If $m$ is odd: Peel off $\sec x \tan x$, use $\sec^2 x = 1 + \tan^2 x$
Type 3: $\int \sin mx \cos nx \, dx$
- Use product-to-sum formulas
Type 4: $\int \frac{1}{a + b\sin x}$ or $\int \frac{1}{a + b\cos x}$
- Use Weierstrass substitution: $t = \tan(x/2)$
Interactive Demo: Trigonometric Integration Patterns
Visualize how trigonometric functions integrate using different techniques.
Type 1: Powers of Sine and Cosine
Case 1a: Odd Power of Sine
Problem: Evaluate $\int \sin^3 x \, dx$
Solution:
Strategy: Save one $\sin x$ for $du$, convert the rest using $\sin^2 x = 1 - \cos^2 x$
$$\sin^3 x = \sin^2 x \cdot \sin x = (1 - \cos^2 x) \sin x$$Let $u = \cos x$, then $du = -\sin x \, dx$
$$\int \sin^3 x \, dx = \int (1 - \cos^2 x) \sin x \, dx$$ $$= \int (1 - u^2)(-du) = -\int (1 - u^2) \, du$$ $$= -\left(u - \frac{u^3}{3}\right) + C = -\cos x + \frac{\cos^3 x}{3} + C$$Verification: $\frac{d}{dx}\left(-\cos x + \frac{\cos^3 x}{3}\right) = \sin x + \cos^2 x \cdot (-\sin x) = \sin x(1 - \cos^2 x) = \sin^3 x$ ✓
Case 1b: Odd Power of Cosine
Problem: Find $\int \sin^2 x \cos^3 x \, dx$
Solution:
Strategy: Save one $\cos x$ for $du$, convert the rest
$$\cos^3 x = \cos^2 x \cdot \cos x = (1 - \sin^2 x) \cos x$$Let $u = \sin x$, then $du = \cos x \, dx$
$$\int \sin^2 x \cos^3 x \, dx = \int \sin^2 x (1 - \sin^2 x) \cos x \, dx$$ $$= \int u^2(1 - u^2) \, du = \int (u^2 - u^4) \, du$$ $$= \frac{u^3}{3} - \frac{u^5}{5} + C = \frac{\sin^3 x}{3} - \frac{\sin^5 x}{5} + C$$Case 1c: Both Powers Even
Problem: Evaluate $\int \sin^2 x \cos^2 x \, dx$
Solution:
Strategy: Use power-reduction formulas repeatedly
Method 1 (Product identity):
$$\sin^2 x \cos^2 x = (\sin x \cos x)^2 = \left(\frac{\sin 2x}{2}\right)^2 = \frac{\sin^2 2x}{4}$$Now use $\sin^2 2x = \frac{1 - \cos 4x}{2}$:
$$\int \sin^2 x \cos^2 x \, dx = \int \frac{1}{4} \cdot \frac{1 - \cos 4x}{2} \, dx$$ $$= \frac{1}{8}\int (1 - \cos 4x) \, dx = \frac{1}{8}\left(x - \frac{\sin 4x}{4}\right) + C$$ $$= \frac{x}{8} - \frac{\sin 4x}{32} + C$$Method 2 (Direct reduction):
$$\sin^2 x = \frac{1 - \cos 2x}{2}, \quad \cos^2 x = \frac{1 + \cos 2x}{2}$$ $$\sin^2 x \cos^2 x = \frac{(1 - \cos 2x)(1 + \cos 2x)}{4} = \frac{1 - \cos^2 2x}{4}$$ $$= \frac{1}{4} - \frac{\cos^2 2x}{4} = \frac{1}{4} - \frac{1}{4} \cdot \frac{1 + \cos 4x}{2}$$ $$= \frac{1}{4} - \frac{1}{8} - \frac{\cos 4x}{8} = \frac{1}{8} - \frac{\cos 4x}{8}$$Both methods give the same answer!
Type 2: Powers of Tangent and Secant
Case 2a: Even Power of Secant
Problem: Evaluate $\int \tan^2 x \sec^4 x \, dx$
Solution:
Strategy: Peel off $\sec^2 x$ (for $du$), convert remaining $\sec^2 x$ using $\sec^2 x = 1 + \tan^2 x$
$$\int \tan^2 x \sec^4 x \, dx = \int \tan^2 x \sec^2 x \cdot \sec^2 x \, dx$$ $$= \int \tan^2 x (1 + \tan^2 x) \sec^2 x \, dx$$Let $u = \tan x$, then $du = \sec^2 x \, dx$
$$= \int u^2(1 + u^2) \, du = \int (u^2 + u^4) \, du$$ $$= \frac{u^3}{3} + \frac{u^5}{5} + C = \frac{\tan^3 x}{3} + \frac{\tan^5 x}{5} + C$$Case 2b: Odd Power of Tangent
Problem: Find $\int \tan^3 x \sec^3 x \, dx$
Solution:
Strategy: Peel off $\sec x \tan x$ (for $du$), convert $\tan^2 x$ using $\tan^2 x = \sec^2 x - 1$
$$\int \tan^3 x \sec^3 x \, dx = \int \tan^2 x \sec^2 x \cdot (\sec x \tan x) \, dx$$ $$= \int (\sec^2 x - 1) \sec^2 x \cdot (\sec x \tan x) \, dx$$Let $u = \sec x$, then $du = \sec x \tan x \, dx$
$$= \int (u^2 - 1) u^2 \, du = \int (u^4 - u^2) \, du$$ $$= \frac{u^5}{5} - \frac{u^3}{3} + C = \frac{\sec^5 x}{5} - \frac{\sec^3 x}{3} + C$$Special Case: Just Tangent or Secant
Problem: Evaluate $\int \tan^4 x \, dx$
Solution:
Strategy: Use $\tan^2 x = \sec^2 x - 1$ repeatedly
$$\tan^4 x = (\tan^2 x)^2 = (\sec^2 x - 1)^2 = \sec^4 x - 2\sec^2 x + 1$$Actually simpler:
$$\tan^4 x = \tan^2 x \cdot \tan^2 x = \tan^2 x(\sec^2 x - 1) = \tan^2 x \sec^2 x - \tan^2 x$$ $$= \tan^2 x \sec^2 x - (\sec^2 x - 1) = \tan^2 x \sec^2 x - \sec^2 x + 1$$ $$\int \tan^4 x \, dx = \int \tan^2 x \sec^2 x \, dx - \int \sec^2 x \, dx + \int 1 \, dx$$First integral: Let $u = \tan x$, $du = \sec^2 x \, dx$
$$\int \tan^2 x \sec^2 x \, dx = \int u^2 \, du = \frac{u^3}{3} = \frac{\tan^3 x}{3}$$ $$\int \tan^4 x \, dx = \frac{\tan^3 x}{3} - \tan x + x + C$$Type 3: Products of Different Angles
Problem: Evaluate $\int \sin 5x \cos 3x \, dx$
Solution:
Use the product-to-sum formula:
$$\sin A \cos B = \frac{1}{2}[\sin(A+B) + \sin(A-B)]$$ $$\sin 5x \cos 3x = \frac{1}{2}[\sin 8x + \sin 2x]$$ $$\int \sin 5x \cos 3x \, dx = \frac{1}{2}\int (\sin 8x + \sin 2x) \, dx$$ $$= \frac{1}{2}\left(-\frac{\cos 8x}{8} - \frac{\cos 2x}{2}\right) + C$$ $$= -\frac{\cos 8x}{16} - \frac{\cos 2x}{4} + C$$Type 4: Rational Functions of Sine and Cosine
Weierstrass Substitution (Universal Substitution)
For integrals of the form $\int R(\sin x, \cos x) \, dx$ where $R$ is a rational function:
Let $t = \tan\left(\frac{x}{2}\right)$, then:
$$\sin x = \frac{2t}{1 + t^2}$$ $$\cos x = \frac{1 - t^2}{1 + t^2}$$ $$dx = \frac{2}{1 + t^2} \, dt$$This converts any trig integral into a rational function in $t$!
Problem: Find $\int \frac{1}{1 + \sin x} \, dx$
Solution:
Let $t = \tan(x/2)$, then $\sin x = \frac{2t}{1+t^2}$ and $dx = \frac{2}{1+t^2} dt$
$$\int \frac{1}{1 + \sin x} \, dx = \int \frac{1}{1 + \frac{2t}{1+t^2}} \cdot \frac{2}{1+t^2} \, dt$$ $$= \int \frac{1}{\frac{1+t^2+2t}{1+t^2}} \cdot \frac{2}{1+t^2} \, dt = \int \frac{2}{(1+t)^2} \, dt$$ $$= \frac{2 \cdot (1+t)^{-1}}{-1} + C = -\frac{2}{1+t} + C$$Substitute back: $t = \tan(x/2)$
$$= -\frac{2}{1 + \tan(x/2)} + C$$Alternative simpler form: Multiply numerator and denominator by $(1 - \sin x)$:
$$\frac{1}{1 + \sin x} = \frac{1 - \sin x}{(1 + \sin x)(1 - \sin x)} = \frac{1 - \sin x}{1 - \sin^2 x} = \frac{1 - \sin x}{\cos^2 x}$$ $$= \sec^2 x - \frac{\sin x}{\cos^2 x} = \sec^2 x - \sec x \tan x$$ $$\int \frac{1}{1 + \sin x} \, dx = \int (\sec^2 x - \sec x \tan x) \, dx$$ $$= \tan x - \sec x + C$$JEE Tip: For specific forms like $\frac{1}{1 \pm \sin x}$ or $\frac{1}{1 \pm \cos x}$, rationalization is faster than Weierstrass!
Special Integrals (Must Remember)
Derivation for $\int \sec x \, dx$:
Multiply by $\frac{\sec x + \tan x}{\sec x + \tan x}$:
$$\int \sec x \, dx = \int \sec x \cdot \frac{\sec x + \tan x}{\sec x + \tan x} \, dx$$ $$= \int \frac{\sec^2 x + \sec x \tan x}{\sec x + \tan x} \, dx$$Let $u = \sec x + \tan x$, then $du = (\sec x \tan x + \sec^2 x) \, dx$
$$= \int \frac{1}{u} \, du = \ln|u| + C = \ln|\sec x + \tan x| + C$$Memory trick: “Sec Plus Tan” (add them in the log)
Advanced Patterns & Tricks
Reduction Formulas
For $I_n = \int \sin^n x \, dx$:
$$I_n = -\frac{\sin^{n-1} x \cos x}{n} + \frac{n-1}{n} I_{n-2}$$Problem: Find $\int \sin^5 x \, dx$
Solution:
Method 1 (Direct): As shown earlier, peel off one $\sin x$
Method 2 (Reduction formula):
$$I_5 = -\frac{\sin^4 x \cos x}{5} + \frac{4}{5} I_3$$ $$I_3 = -\frac{\sin^2 x \cos x}{3} + \frac{2}{3} I_1$$ $$I_1 = \int \sin x \, dx = -\cos x + C$$Work backwards:
$$I_3 = -\frac{\sin^2 x \cos x}{3} + \frac{2}{3}(-\cos x) = -\frac{\sin^2 x \cos x}{3} - \frac{2\cos x}{3}$$ $$I_5 = -\frac{\sin^4 x \cos x}{5} + \frac{4}{5}\left(-\frac{\sin^2 x \cos x}{3} - \frac{2\cos x}{3}\right)$$ $$= -\frac{\sin^4 x \cos x}{5} - \frac{4\sin^2 x \cos x}{15} - \frac{8\cos x}{15} + C$$For JEE: Direct method is usually faster unless you need a general formula!
Memory Tricks & Patterns
Quick Reference Guide
| Integral | Strategy | Key Identity |
|---|---|---|
| $\int \sin^m x \cos^n x \, dx$ (odd power) | Substitution | Save one factor for $du$ |
| $\int \sin^m x \cos^n x \, dx$ (both even) | Power reduction | $\sin^2 x = \frac{1-\cos 2x}{2}$ |
| $\int \tan^m x \sec^n x \, dx$ ($n$ even) | Peel $\sec^2 x$ | $\sec^2 x = 1 + \tan^2 x$ |
| $\int \tan^m x \sec^n x \, dx$ ($m$ odd) | Peel $\sec x \tan x$ | $\tan^2 x = \sec^2 x - 1$ |
| $\int \sin Ax \cos Bx \, dx$ | Product-to-sum | Convert to sum of sines |
| $\int \frac{1}{a + b\sin x} \, dx$ | Weierstrass or rationalize | $t = \tan(x/2)$ |
The “Save and Convert” Mnemonic
"Save One Convert Rest" = SOCR
For $\int \sin^m x \cos^n x \, dx$:
- Save one factor of the odd power
- Convert the rest using Pythagorean identity
- Replace with substitution
Common Mistakes to Avoid
WRONG: $\int \sin^2 x \, dx = \int (1 - \cos^2 x) \, dx$ (doesn’t help!)
CORRECT: Use power reduction: $\sin^2 x = \frac{1 - \cos 2x}{2}$
Why? The substitution $\sin^2 x = 1 - \cos^2 x$ doesn’t reduce the power—you still have $\cos^2 x$ to integrate!
Problem: $\int \sin^3 x \, dx$ with $u = \cos x$
WRONG: $du = \cos x \, dx$ (missing negative!)
CORRECT: $du = -\sin x \, dx$, so $\sin x \, dx = -du$
Always check the sign when substituting!
WRONG: $\sin A \cos B = \sin(A+B) + \sin(A-B)$
CORRECT: $\sin A \cos B = \frac{1}{2}[\sin(A+B) + \sin(A-B)]$
That $\frac{1}{2}$ is crucial!
Practice Problems
Level 1: Foundation (NCERT)
Evaluate: $\int \cos^2 x \, dx$
Solution:
$$\cos^2 x = \frac{1 + \cos 2x}{2}$$ $$\int \cos^2 x \, dx = \int \frac{1 + \cos 2x}{2} \, dx = \frac{1}{2}\left(x + \frac{\sin 2x}{2}\right) + C$$ $$= \frac{x}{2} + \frac{\sin 2x}{4} + C$$Find: $\int \sin x \cos x \, dx$
Solution:
Method 1 (Substitution): Let $u = \sin x$, $du = \cos x \, dx$
$$\int \sin x \cos x \, dx = \int u \, du = \frac{u^2}{2} + C = \frac{\sin^2 x}{2} + C$$Method 2 (Identity): $\sin x \cos x = \frac{\sin 2x}{2}$
$$\int \sin x \cos x \, dx = \int \frac{\sin 2x}{2} \, dx = -\frac{\cos 2x}{4} + C$$Both are correct! (They differ by a constant)
Level 2: JEE Main
JEE Main Pattern: Evaluate $\int \sin^4 x \, dx$
Solution:
Use power reduction twice:
$$\sin^2 x = \frac{1 - \cos 2x}{2}$$ $$\sin^4 x = (\sin^2 x)^2 = \left(\frac{1 - \cos 2x}{2}\right)^2 = \frac{1 - 2\cos 2x + \cos^2 2x}{4}$$For $\cos^2 2x$, use: $\cos^2 2x = \frac{1 + \cos 4x}{2}$
$$\sin^4 x = \frac{1 - 2\cos 2x}{4} + \frac{1 + \cos 4x}{8}$$ $$= \frac{2 - 4\cos 2x + 1 + \cos 4x}{8} = \frac{3 - 4\cos 2x + \cos 4x}{8}$$ $$\int \sin^4 x \, dx = \frac{1}{8}\int (3 - 4\cos 2x + \cos 4x) \, dx$$ $$= \frac{1}{8}\left(3x - 4 \cdot \frac{\sin 2x}{2} + \frac{\sin 4x}{4}\right) + C$$ $$= \frac{3x}{8} - \frac{\sin 2x}{4} + \frac{\sin 4x}{32} + C$$Problem: Find $\int \tan^3 x \, dx$
Solution:
$$\tan^3 x = \tan x \cdot \tan^2 x = \tan x(\sec^2 x - 1) = \tan x \sec^2 x - \tan x$$ $$\int \tan^3 x \, dx = \int \tan x \sec^2 x \, dx - \int \tan x \, dx$$First integral: Let $u = \tan x$, $du = \sec^2 x \, dx$
$$\int \tan x \sec^2 x \, dx = \int u \, du = \frac{u^2}{2} = \frac{\tan^2 x}{2}$$Second integral:
$$\int \tan x \, dx = \int \frac{\sin x}{\cos x} \, dx = -\ln|\cos x| + C = \ln|\sec x| + C$$ $$\int \tan^3 x \, dx = \frac{\tan^2 x}{2} - \ln|\sec x| + C$$Problem: Evaluate $\int \frac{1}{1 + \cos x} \, dx$
Solution:
Multiply by $\frac{1 - \cos x}{1 - \cos x}$:
$$\frac{1}{1 + \cos x} = \frac{1 - \cos x}{(1 + \cos x)(1 - \cos x)} = \frac{1 - \cos x}{1 - \cos^2 x} = \frac{1 - \cos x}{\sin^2 x}$$ $$= \frac{1}{\sin^2 x} - \frac{\cos x}{\sin^2 x} = \csc^2 x - \cot x \csc x$$ $$\int \frac{1}{1 + \cos x} \, dx = \int (\csc^2 x - \cot x \csc x) \, dx$$ $$= -\cot x - (-\csc x) + C = -\cot x + \csc x + C$$Alternative: Use half-angle: $1 + \cos x = 2\cos^2(x/2)$
$$\int \frac{1}{1 + \cos x} \, dx = \int \frac{1}{2\cos^2(x/2)} \, dx = \frac{1}{2}\int \sec^2(x/2) \, dx = \tan(x/2) + C$$Level 3: JEE Advanced
JEE Advanced Pattern: Evaluate $\int \frac{\sin x}{\sin 3x} \, dx$
Solution:
Use $\sin 3x = 3\sin x - 4\sin^3 x = \sin x(3 - 4\sin^2 x)$
$$\frac{\sin x}{\sin 3x} = \frac{\sin x}{\sin x(3 - 4\sin^2 x)} = \frac{1}{3 - 4\sin^2 x}$$Use $\sin^2 x = \frac{1 - \cos 2x}{2}$:
$$3 - 4\sin^2 x = 3 - 4 \cdot \frac{1 - \cos 2x}{2} = 3 - 2(1 - \cos 2x) = 1 + 2\cos 2x$$ $$\int \frac{\sin x}{\sin 3x} \, dx = \int \frac{1}{1 + 2\cos 2x} \, dx$$This requires Weierstrass or a clever transformation. For JEE, recognize this as a standard form!
Alternative approach: This is actually best solved using partial fractions after converting to exponential form (beyond JEE scope).
Key learning: Some trig integrals are extremely difficult—in JEE, if you don’t see a pattern in 30 seconds, move on!
Conceptual: If $I = \int \sin^6 x \, dx$ and $J = \int \cos^6 x \, dx$, prove that $I + J = \frac{5x}{16} + \text{(trig terms)}$
Solution:
Use the identity: $(\sin^2 x + \cos^2 x)^3 = 1$
Expand: $\sin^6 x + 3\sin^4 x \cos^2 x + 3\sin^2 x \cos^4 x + \cos^6 x = 1$
So: $\sin^6 x + \cos^6 x = 1 - 3\sin^2 x \cos^2 x(sin^2 x + \cos^2 x)$
$$= 1 - 3\sin^2 x \cos^2 x$$We know: $\sin^2 x \cos^2 x = \frac{\sin^2 2x}{4} = \frac{1}{4} \cdot \frac{1 - \cos 4x}{2} = \frac{1 - \cos 4x}{8}$
$$\sin^6 x + \cos^6 x = 1 - 3 \cdot \frac{1 - \cos 4x}{8} = 1 - \frac{3}{8} + \frac{3\cos 4x}{8} = \frac{5}{8} + \frac{3\cos 4x}{8}$$ $$I + J = \int \left(\frac{5}{8} + \frac{3\cos 4x}{8}\right) dx = \frac{5x}{8} + \frac{3\sin 4x}{32} + C$$Wait, the problem says $\frac{5x}{16}$—let me recalculate…
Actually, the coefficient should be $\frac{5}{8}$. The problem might have a typo or uses a different identity.
Key skill: For JEE Advanced, knowing algebraic identities is as important as integration techniques!
Quick Revision Box
| Type | Example | First Step |
|---|---|---|
| Odd power (sin/cos) | $\int \sin^3 x \, dx$ | Save one, substitute $u = \cos x$ |
| Both even | $\int \sin^2 x \cos^2 x \, dx$ | Power reduction formulas |
| Even secant | $\int \sec^4 x \, dx$ | Peel $\sec^2 x$, use $\sec^2 x = 1 + \tan^2 x$ |
| Odd tangent | $\int \tan^3 x \sec x \, dx$ | Peel $\sec x \tan x$, sub $u = \sec x$ |
| Product angles | $\int \sin 3x \cos 5x \, dx$ | Product-to-sum formula |
| $\frac{1}{1+\sin x}$ | $\int \frac{1}{1+\sin x} \, dx$ | Rationalize (multiply by $1-\sin x$) |
Cross-Links & Prerequisites
Before studying this topic:
- Indefinite Integrals - Basic integration
- Integration Techniques - Substitution method
- Trigonometry - Trig identities
Related topics:
- Definite Integrals - Apply these techniques with limits
- Applications of Integrals - Find areas under trig curves
Teacher’s Summary
Know Your Identities: Trig integration is 70% knowing which identity to use. Master power-reduction and product-to-sum formulas.
Odd Powers = Substitution: If either sine or cosine has an odd power, save one factor for $du$ and substitute. This is the most common JEE pattern.
Even Powers = Reduction: When both powers are even, use $\sin^2 x = \frac{1-\cos 2x}{2}$ and $\cos^2 x = \frac{1+\cos 2x}{2}$ to reduce the power.
SOCR for Mixed Powers: Save One, Convert Rest. This simple rule solves 80% of trig integration problems.
Special Forms Need Special Tricks: For $\int \sec x \, dx$ and $\int \csc x \, dx$, just memorize the results. For rational functions in sin/cos, try rationalization before using Weierstrass.
“Trigonometric integration is pattern recognition. The more problems you solve, the faster you’ll spot the right identity. Aim for 50+ practice problems to master this topic.”
JEE Strategy: Most JEE problems test 2-3 standard patterns repeatedly. Focus on mastering $\int \sin^m x \cos^n x \, dx$ and $\int \tan^m x \sec^n x \, dx$—these cover 80% of questions.
What’s Next?
You’ve conquered trigonometric integration! Now apply these skills:
- Next: Definite Integrals - Add limits and use properties
- Then: Applications of Integrals - Calculate areas under trig curves
- Advanced: Differential Equations - Use trig integration to solve ODEs
Keep practicing—trig integration is like learning a musical instrument. The first 20 problems are hard, the next 30 get easier, and after 50, it becomes intuitive!