Advanced Integration Techniques

Module 2 Objective

This module covers three high-yield JEE techniques: Integration by Parts (IBP), Partial Fractions, and Special Trigonometric Integrals. Together, these methods solve 90% of the complex integration questions that appear in JEE Main and Advanced.

Part A: Integration by Parts (IBP)

IBP is used when the integrand is a product of two fundamentally different types of functions (e.g., polynomial × trig, polynomial × log, \(e^x\) × trig).

The IBP Formula \[\int u \cdot v\,dx = u\int v\,dx - \int\!\left(\frac{du}{dx}\cdot\int v\,dx\right)dx\]

ILATE Rule — Pick \(u\) (to differentiate) in this descending priority:
Inverse Trig → Logarithm → Algebraic (polynomial) → Trigonometric → Exponential
The remaining factor becomes \(v\) (to integrate).

When to Use Each Combination

Integrand Type	Choose \(u\)	Choose \(v\,dx\)
\(x^n \ln x\)	\(\ln x\) (L before A)	\(x^n\,dx\)
\(x^n \sin x\) or \(x^n \cos x\)	\(x^n\) (A before T)	\(\sin x\,dx\) or \(\cos x\,dx\)
\(x^n e^x\)	\(x^n\) (A before E)	\(e^x\,dx\)
\(e^{ax}\sin bx\) or \(e^{ax}\cos bx\)	Either (cyclical — apply IBP twice)	The other
\(\sin^{-1}x\) or \(\tan^{-1}x\) alone	The inverse trig (I)	\(dx\) → \(\int 1\,dx = x\)

Q1. Evaluate \(\displaystyle\int x\cos x\,dx\) Solution — ILATE: \(u=x\) (A), \(v=\cos x\) \[\int x\cos x\,dx = x(\sin x) - \int 1\cdot\sin x\,dx = x\sin x+\cos x+C\]

Q2. Evaluate \(\displaystyle\int x^2 e^x\,dx\) Solution — Apply IBP twice: Round 1: \(u=x^2\), \(v=e^x\): \(= x^2 e^x - \displaystyle\int 2xe^x\,dx\)
Round 2 on \(\int 2xe^x\): \(u=2x\), \(v=e^x\): \(= 2xe^x - 2e^x\)
\[\boxed{\int x^2 e^x\,dx = e^x(x^2-2x+2)+C}\]

Q3. Evaluate \(\displaystyle\int \ln x\,dx\) Solution — \(u=\ln x\) (L), \(v=1\) so \(\int v\,dx = x\) \[\int \ln x \cdot 1\,dx = x\ln x - \int \frac{1}{x}\cdot x\,dx = x\ln x - x + C = \boxed{x(\ln x - 1)+C}\]

Q4. Evaluate \(\displaystyle\int e^{2x}\sin 3x\,dx\) Solution — Cyclical IBP: Let \(I = \int e^{2x}\sin 3x\,dx\). IBP with \(u=\sin 3x\), \(v=e^{2x}\):
\(I = \dfrac{e^{2x}\sin 3x}{2} - \dfrac{3}{2}\int e^{2x}\cos 3x\,dx\)
IBP again on the new integral: \(\int e^{2x}\cos 3x\,dx = \dfrac{e^{2x}\cos 3x}{2}+\dfrac{3}{2}\int e^{2x}\sin 3x\,dx\)
Substituting back: \(I = \dfrac{e^{2x}\sin 3x}{2} - \dfrac{3}{2}\left[\dfrac{e^{2x}\cos 3x}{2}+\dfrac{3}{2}I\right]\)
\(I + \dfrac{9}{4}I = \dfrac{e^{2x}(2\sin 3x-3\cos 3x)}{4} \Rightarrow \dfrac{13I}{4} = \dfrac{e^{2x}(2\sin 3x-3\cos 3x)}{4}\)
\[\boxed{I = \frac{e^{2x}(2\sin 3x-3\cos 3x)}{13}+C}\]

JEE Star Formula — Memorise This! \[\int e^x[f(x)+f'(x)]\,dx = e^x\cdot f(x) + C\]

This form appears in almost every JEE Main paper. The strategy: spot \(f(x)\) in the first bracket and confirm the second bracket equals \(f'(x)\). Examples:
\(\int e^x(\sin x+\cos x)\,dx = e^x\sin x+C\) | \(\int e^x\left(\dfrac{1}{x}-\dfrac{1}{x^2}\right)dx = \dfrac{e^x}{x}+C\) | \(\int e^x(\tan x+\sec^2 x)\,dx = e^x\tan x+C\)

Q5. Evaluate \(\displaystyle\int e^x\left(\frac{x-1}{x^2}\right)dx\) Solution — Identify \(f(x)\): Write integrand as \(e^x\left[\dfrac{1}{x}-\dfrac{1}{x^2}\right]\). Here \(f(x)=\dfrac{1}{x}\), \(f'(x)=-\dfrac{1}{x^2}\). ✓
\(\boxed{= \dfrac{e^x}{x}+C}\)

Q6. Evaluate \(\displaystyle\int \tan^{-1}x\,dx\) Solution — \(u=\tan^{-1}x\) (I), \(\int v\,dx = x\) \[\int \tan^{-1}x\,dx = x\tan^{-1}x - \int\frac{x}{1+x^2}\,dx = x\tan^{-1}x - \frac{1}{2}\ln(1+x^2)+C\]

Part B: Partial Fractions

Used to integrate rational functions \(\dfrac{P(x)}{Q(x)}\). If deg\((P) \ge\) deg\((Q)\), first perform long division.

Complete Decomposition Guide

Factor in Q(x)	Partial Fraction Form
Linear: \((x-a)\)	\(\dfrac{A}{x-a}\)
Repeated Linear: \((x-a)^2\)	\(\dfrac{A}{x-a}+\dfrac{B}{(x-a)^2}\)
Repeated Linear: \((x-a)^n\)	\(\dfrac{A_1}{x-a}+\dfrac{A_2}{(x-a)^2}+\cdots+\dfrac{A_n}{(x-a)^n}\)
Irreducible Quadratic: \((x^2+bx+c)\)	\(\dfrac{Ax+B}{x^2+bx+c}\)

Finding Constants — Two Methods

Substitution Method (Heaviside Cover-up)
Substitute the root of each linear factor to quickly get that constant. E.g., for factor \((x-2)\), put \(x=2\).

Comparison of Coefficients
Expand the right side, then equate coefficients of like powers of \(x\) to form a system of equations for the remaining constants.

Q7. Evaluate \(\displaystyle\int\frac{2x+5}{(x-1)(x+2)}\,dx\) Solution: Let \(\dfrac{2x+5}{(x-1)(x+2)} = \dfrac{A}{x-1}+\dfrac{B}{x+2}\). Multiply through:
\(2x+5 = A(x+2)+B(x-1)\)
\(x=1\): \(7=3A \Rightarrow A=7/3\).
\(x=-2\): \(1=-3B \Rightarrow B=-1/3\).
\(\displaystyle\int\left[\frac{7/3}{x-1}+\frac{-1/3}{x+2}\right]dx = \boxed{\frac{7}{3}\ln|x-1|-\frac{1}{3}\ln|x+2|+C}\)

Q8. Evaluate \(\displaystyle\int\frac{x^2+1}{x(x-1)^2}\,dx\) Solution: \(\dfrac{x^2+1}{x(x-1)^2} = \dfrac{A}{x}+\dfrac{B}{x-1}+\dfrac{C}{(x-1)^2}\)
\(x=0\): \(1=A\). \(x=1\): \(2=C\). Coefficient of \(x^2\): \(1=A+B \Rightarrow B=0\).
\(\displaystyle\int\left[\frac{1}{x}+\frac{2}{(x-1)^2}\right]dx = \ln|x|-\frac{2}{x-1}+C\)

Q9. Evaluate \(\displaystyle\int\frac{3x+1}{(x-2)(x^2+1)}\,dx\) Solution: \(\dfrac{3x+1}{(x-2)(x^2+1)} = \dfrac{A}{x-2}+\dfrac{Bx+C}{x^2+1}\)
\(x=2\): \(7=5A \Rightarrow A=7/5\).
Expanding and comparing:\(x^2\): \(0=A+B \Rightarrow B=-7/5\). Constant: \(1=-2A+C-2B \Rightarrow C=1/5\).
\(\displaystyle\int\!\left[\frac{7/5}{x-2}+\frac{-7x/5+1/5}{x^2+1}\right]dx = \frac{7}{5}\ln|x-2|-\frac{7}{10}\ln(x^2+1)+\frac{1}{5}\tan^{-1}x+C\)

Q10. Evaluate \(\displaystyle\int\frac{x^3+x+1}{x^2-1}\,dx\) Solution — Long Division First (deg of num ≥ deg of denom): \(x^3+x+1 \div (x^2-1)\): Quotient \(= x\), Remainder \(= 2x+1\).
\(\displaystyle\int x\,dx + \int\frac{2x+1}{x^2-1}\,dx = \frac{x^2}{2}+\int\frac{2x+1}{(x-1)(x+1)}\,dx\)
Partial fractions: \(\dfrac{2x+1}{(x-1)(x+1)} = \dfrac{3/2}{x-1}+\dfrac{1/2}{x+1}\)
\(= \dfrac{x^2}{2}+\dfrac{3}{2}\ln|x-1|+\dfrac{1}{2}\ln|x+1|+C\)

Part C: Special Trigonometric Integrals

Type 1 — Powers of Sin/Cos

Core Identities

\(\sin^2 x = \dfrac{1-\cos 2x}{2}\) \(\cos^2 x = \dfrac{1+\cos 2x}{2}\)
\(\sin^3 x = \dfrac{3\sin x-\sin 3x}{4}\) \(\cos^3 x = \dfrac{3\cos x+\cos 3x}{4}\)
\(\sin x\cos x = \dfrac{\sin 2x}{2}\)
\(\sin^2 x\cos^2 x = \dfrac{1-\cos 4x}{8}\)

Q11. Evaluate \(\displaystyle\int\sin^4 x\,dx\) Solution: \(\sin^4 x = \left(\dfrac{1-\cos 2x}{2}\right)^2 = \dfrac{1-2\cos 2x+\cos^2 2x}{4} = \dfrac{1-2\cos 2x+\frac{1+\cos 4x}{2}}{4}\)
\(= \dfrac{3}{8}-\dfrac{\cos 2x}{2}+\dfrac{\cos 4x}{8}\)
\(\displaystyle\int\sin^4 x\,dx = \frac{3x}{8}-\frac{\sin 2x}{4}+\frac{\sin 4x}{32}+C\)

Type 2 — Integrals of the form \(\int \frac{dx}{a+b\sin x}\), \(\int\frac{dx}{a+b\cos x}\)

Weierstrass / t-Substitution

Put \(t = \tan\dfrac{x}{2}\). Then:
\(\sin x = \dfrac{2t}{1+t^2}\) \(\cos x = \dfrac{1-t^2}{1+t^2}\) \(dx = \dfrac{2\,dt}{1+t^2}\)
This converts any rational function of \(\sin x, \cos x\) into a rational function of \(t\).

Q12. Evaluate \(\displaystyle\int\frac{dx}{3+2\cos x}\) Solution using \(t=\tan(x/2)\): \(\cos x = \dfrac{1-t^2}{1+t^2}\), \(dx = \dfrac{2dt}{1+t^2}\).
\(\displaystyle\int\frac{1}{3+2\cdot\frac{1-t^2}{1+t^2}}\cdot\frac{2\,dt}{1+t^2} = \int\frac{2\,dt}{3(1+t^2)+2(1-t^2)} = \int\frac{2\,dt}{5+t^2}\)
\(= \dfrac{2}{\sqrt{5}}\tan^{-1}\!\dfrac{t}{\sqrt{5}}+C = \boxed{\dfrac{2}{\sqrt{5}}\tan^{-1}\!\left(\dfrac{\tan(x/2)}{\sqrt{5}}\right)+C}\)

Type 3 — \(\int\frac{dx}{a\sin^2 x + b\cos^2 x}\) type

Technique

Divide numerator and denominator by \(\cos^2 x\). The denominator becomes \(a\tan^2x + b\). Let \(t = \tan x\), \(dt = \sec^2 x\,dx\). This directly reduces to \(\int\frac{dt}{at^2+b}\) — a standard inverse tan form.

Q13. Evaluate \(\displaystyle\int\frac{dx}{4\sin^2 x+9\cos^2 x}\) Solution: Divide by \(\cos^2x\): \(\displaystyle\int\frac{\sec^2 x\,dx}{4\tan^2 x+9}\)
Let \(t=\tan x\), \(dt=\sec^2 x\,dx\):
\(\displaystyle\int\frac{dt}{4t^2+9} = \frac{1}{4}\int\frac{dt}{t^2+(3/2)^2} = \frac{1}{4}\cdot\frac{1}{3/2}\tan^{-1}\!\frac{2t}{3}+C = \boxed{\frac{1}{6}\tan^{-1}\!\frac{2\tan x}{3}+C}\)

Type 4 — \(\int\frac{p\sin x+q\cos x}{a\sin x+b\cos x}\,dx\) type

Technique

Express numerator as \(\lambda(\text{denominator})+\mu(\text{derivative of denominator})\). Find \(\lambda, \mu\) by comparing coefficients. The integral splits into \(\lambda x + \mu\ln|\text{denominator}|+C\).

Q14. Evaluate \(\displaystyle\int\frac{3\sin x+2\cos x}{2\sin x+3\cos x}\,dx\) Solution: Write \(3\sin x+2\cos x = \lambda(2\sin x+3\cos x)+\mu(2\cos x-3\sin x)\)
Comparing \(\sin x\): \(3=2\lambda-3\mu\). Comparing \(\cos x\): \(2=3\lambda+2\mu\).
Solving: \(\lambda=12/13\), \(\mu=-5/13\).
\(\displaystyle\int\left[\frac{12}{13}+\frac{-5}{13}\cdot\frac{2\cos x-3\sin x}{2\sin x+3\cos x}\right]dx = \frac{12x}{13}-\frac{5}{13}\ln|2\sin x+3\cos x|+C\)

Type 5 — \(\int\frac{px+q}{ax^2+bx+c}\,dx\) and \(\int\frac{px+q}{\sqrt{ax^2+bx+c}}\,dx\)

Technique

Express numerator \(px+q = \lambda(2ax+b)+\mu\). Find \(\lambda = p/(2a)\) and \(\mu = q-pb/(2a)\). The first part gives a log/root, the second gives an inverse trig after completing the square.

Q15. Evaluate \(\displaystyle\int\frac{x+3}{x^2+2x+5}\,dx\) Solution: Write \(x+3 = \lambda(2x+2)+\mu\). Solving: \(\lambda=1/2\), \(\mu=2\).
\(\displaystyle\int\!\left[\frac{1}{2}\cdot\frac{2x+2}{x^2+2x+5}+\frac{2}{x^2+2x+5}\right]dx = \frac{1}{2}\ln(x^2+2x+5)+2\int\frac{dx}{(x+1)^2+4}\)
\(\displaystyle = \frac{1}{2}\ln(x^2+2x+5)+\tan^{-1}\!\frac{x+1}{2}+C\)

Q16. Evaluate \(\displaystyle\int\frac{2x+3}{\sqrt{x^2+4x+1}}\,dx\) Solution: Write \(2x+3 = \lambda(2x+4)+\mu\). Solving: \(\lambda=1\), \(\mu=-1\).
\(\displaystyle\int\!\left[\frac{2x+4}{\sqrt{x^2+4x+1}}-\frac{1}{\sqrt{x^2+4x+1}}\right]dx\)
First part: \(2\sqrt{x^2+4x+1}\). Second: complete the square — \(\sqrt{(x+2)^2-3}\), use formula 24.
\(\displaystyle = 2\sqrt{x^2+4x+1}-\ln\!\left|(x+2)+\sqrt{x^2+4x+1}\right|+C\)

Chapter Summary — Decision Flowchart

Integrand Type	Method
Composite function with inner derivative present	Substitution
Product of two different function types	IBP (use ILATE)
Form \(e^x[f(x)+f'(x)]\)	Direct formula: \(e^x f(x)+C\)
Rational function \(P(x)/Q(x)\)	Partial Fractions (divide first if needed)
Rational function of \(\sin x, \cos x\)	Weierstrass t-substitution
\(\dfrac{dx}{a\sin^2x+b\cos^2x}\)	Divide by \(\cos^2x\), then substitute \(t=\tan x\)
\(\dfrac{p\sin x+q\cos x}{a\sin x+b\cos x}\)	Decompose numerator as \(\lambda(\text{denom})+\mu(\text{denom}')\)
\(\dfrac{px+q}{ax^2+bx+c}\)	Decompose numerator, log + inverse trig

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