Methods of Integration

If the integrand can be written as $f(g(x))\cdot g^{\prime }(x)$, substitute $t=g(x)$ so that $dt=g^{\prime }(x)dx$:

${\displaystyle \int f(g(x))\cdot g^{\prime }(x)dx=\int f(t)dt}$

The key skill is spotting the inner function $g(x)$ whose derivative $g^{\prime }(x)$ also appears (or can be adjusted by a constant) in the integrand.

When to Use Substitution

• Integrand contains a composite function $f(g(x))$ and $g^{\prime }(x)$ is also present.
• The form ${\displaystyle \frac{f^{\prime }(x)}{f(x)}}$ → substitute $t=f(x)$, giving $\ln |f(x)|+C$.
• The form $[f(x){]}^n\cdot f^{\prime }(x)$ → substitute $t=f(x)$, giving ${\displaystyle \frac{[f(x){]}^{n+1}}{n+1}}+C$.

Standard Substitutions for Specific Forms

Worked Example 1

${\displaystyle \int {\displaystyle \frac{2x}{x^2+1}}dx}$

Let $t=x^2+1$, so $dt=2xdx$.
$={\displaystyle \int {\displaystyle \frac{dt}{t}}=\ln |t|+C=\ln |x^2+1|+C=\ln (x^2+1)+C}$.

Worked Example 2 — Trigonometric Substitution

${\displaystyle \int \sqrt{4-x^2}dx}$

Let $x=2\sin \theta$, so $dx=2\cos \theta d\theta$ and $\sqrt{4-x^2}=2\cos \theta$.
$={\displaystyle \int 2\cos \theta \cdot 2\cos \theta d\theta =4\int {\cos }^2\theta d\theta }$
$=4\cdot {\displaystyle \frac{\theta }{2}}+4\cdot {\displaystyle \frac{\sin 2\theta }{4}}+C$
$=2\theta +\sin 2\theta +C$
$=2{\sin }^{-1}\left({\displaystyle \frac{x}{2}}\right)+{\displaystyle \frac{x\sqrt{4-x^2}}{2}}+C$.

Special Substitution — $t=\tan (x/2)$ (Weierstrass)

For integrals of the form ${\displaystyle \int {\displaystyle \frac{dx}{a+b\sin x}}}$ or ${\displaystyle \int {\displaystyle \frac{dx}{a+b\cos x}}}$, substitute $t=\tan \left({\displaystyle \frac{x}{2}}\right)$:

$\sin x={\displaystyle \frac{2t}{1+t^2}}$,    $\cos x={\displaystyle \frac{1-t^2}{1+t^2}}$,    $dx={\displaystyle \frac{2dt}{1+t^2}}$

This converts any rational function of $\sin x$ and $\cos x$ into a rational function of $t$.

Integration by Parts is based on the product rule of differentiation. For two functions $u$ and $v$:

${\displaystyle \int u\cdot vdx=u\int vdx-\int (\frac{du}{dx}\cdot \int vdx)dx}$

Or in the compact ILATE notation:

${\displaystyle \int udv=uv-\int vdu}$

ILATE Rule — Choosing $u$ (the Function to Differentiate)

Choose $u$ as the function that appears earlier in the ILATE order:

The function later in ILATE becomes $dv$ (the function to integrate). ILATE is a guideline, not an absolute rule — context sometimes overrides it.

Worked Example 3

${\displaystyle \int x{e}^{x}dx}$

Choose $u=x$ (Algebraic, before Exponential in ILATE) and $dv=e^{x}dx$.
$du=dx$, $v=e^x$.
$=x{e}^x-{\displaystyle \int e^{x}dx=x{e}^x-e^x+C=e^x(x-1)+C}$.

Worked Example 4 — Cyclic Integration

${\displaystyle \int e^x\sin xdx}$

Let $I={\displaystyle \int e^x\sin xdx}$. Apply IBP twice:
Take $u=\sin x$, $dv=e^{x}dx$:
$I=e^x\sin x-{\displaystyle \int e^x\cos xdx}$
Apply IBP again on ${\displaystyle \int e^x\cos xdx}$ with $u=\cos x$:
$I=e^x\sin x-[e^x\cos x+{\displaystyle \int e^x\sin xdx}]$
$I=e^x\sin x-e^x\cos x-I$
$2I=e^x(\sin x-\cos x)$
$I={\displaystyle \frac{e^x(\sin x-\cos x)}{2}}+C$.

Key Results from Integration by Parts

The $e^x[f(x)+f^{\prime }(x)]$ Shortcut — Revisited

This is actually a consequence of Integration by Parts with $u=f(x)$ and $dv=e^{x}dx$:

${\displaystyle \int e^x[f(x)+f^{\prime }(x)]dx=e^{x}f(x)+C}$

Always scan for this pattern before attempting full IBP — it saves significant time in JEE.

When the integrand involves powers or products of trigonometric functions, the strategy is to use identities to reduce the integrand to a sum of standard forms.

Type A — Powers of $\sin x$ and $\cos x$

Type B — Products of $\sin$ and $\cos$ with different angles

Use product-to-sum identities:

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

Type C — Powers of $\tan x$, $\sec x$, $\cot x$, $\csc x$

Worked Example 5

${\displaystyle \int {\sin }^{3}xdx}$

${\sin }^{3}x=\sin x\cdot {\sin }^{2}x=\sin x(1-{\cos }^{2}x)$
$={\displaystyle \int \sin xdx-\int \sin x{\cos }^{2}xdx}$
For second integral, let $t=\cos x$, $dt=-\sin xdx$:
$=-\cos x-{\displaystyle \int t^2(-dt)=-\cos x+{\displaystyle \frac{t^3}{3}}+C}$
$=-\cos x+{\displaystyle \frac{{\cos }^{3}x}{3}}+C$.

These are standard techniques for integrating rational and near-rational functions that reduce to known forms after algebraic manipulation.

Type 1 — ${\displaystyle \int {\displaystyle \frac{dx}{a{x}^2+bx+c}}}$

Step: Complete the square in the denominator, then match to ${\displaystyle \frac{1}{t^2+k^2}}$ or ${\displaystyle \frac{1}{t^2-k^2}}$ form.

Result: ${\tan }^{-1}$ or $\ln \left|{\displaystyle \frac{\cdot }{\cdot }}\right|$ type.

Type 2 — ${\displaystyle \int {\displaystyle \frac{dx}{\sqrt{a{x}^2+bx+c}}}}$

Step: Complete the square, then match to ${\displaystyle \frac{1}{\sqrt{a^2-t^2}}}$, ${\displaystyle \frac{1}{\sqrt{t^2+a^2}}}$, or ${\displaystyle \frac{1}{\sqrt{t^2-a^2}}}$ form.

Result: ${\sin }^{-1}$ or $\ln |t+\sqrt{\cdot }|$ type.

Type 3 — ${\displaystyle \int {\displaystyle \frac{px+q}{a{x}^2+bx+c}}dx}$

Step: Write $px+q=A{\displaystyle \frac{d}{dx}}(a{x}^2+bx+c)+B$
$=A(2ax+b)+B$.
Solve for $A$ and $B$ by comparing coefficients. Then split:

${\displaystyle \int {\displaystyle \frac{px+q}{a{x}^2+bx+c}}dx}$
$=A\int {\displaystyle \frac{2ax+b}{a{x}^2+bx+c}}dx+B\int {\displaystyle \frac{dx}{a{x}^2+bx+c}}$

The first integral gives $\ln |a{x}^2+bx+c|$; the second is Type 1.

Type 4 — ${\displaystyle \int {\displaystyle \frac{px+q}{\sqrt{a{x}^2+bx+c}}}dx}$

Step: Same decomposition as Type 3: write $px+q=A(2ax+b)+B$, then split.
First integral gives $2\sqrt{a{x}^2+bx+c}$; second is Type 2.

Type 5 — ${\displaystyle \int {\displaystyle \frac{dx}{(px+q)\sqrt{a{x}^2+bx+c}}}}$

Step: Substitute $t={\displaystyle \frac{1}{px+q}}$ to eliminate the linear factor.

Worked Example 6

${\displaystyle \int {\displaystyle \frac{x+1}{x^2+2x+3}}dx}$

Write $x+1={\displaystyle \frac{1}{2}}(2x+2)+0$, so $A={\displaystyle \frac{1}{2}}$, $B=0$.
$={\displaystyle \frac{1}{2}}{\displaystyle \int {\displaystyle \frac{2x+2}{x^2+2x+3}}dx}$
$={\displaystyle \frac{1}{2}}\ln |x^2+2x+3|+C$.

Worked Example 7

${\displaystyle \int {\displaystyle \frac{2x+3}{x^2+2x+5}}dx}$

Write $2x+3=A(2x+2)+B$.
Comparing: $A=1$, $B=1$.
$={\displaystyle \int {\displaystyle \frac{2x+2}{x^2+2x+5}}dx+\int {\displaystyle \frac{dx}{x^2+2x+5}}}$
$=\ln |x^2+2x+5|+{\displaystyle \int {\displaystyle \frac{dx}{(x+1{)}^2+4}}}$
$=\ln |x^2+2x+5|+{\displaystyle \frac{1}{2}}{\tan }^{-1}\left({\displaystyle \frac{x+1}{2}}\right)+C$.

When the integrand is a proper rational function ${\displaystyle \frac{P(x)}{Q(x)}}$ (degree of $P$ less than degree of $Q$), decompose it into simpler partial fractions. If degree of $P\ge$ degree of $Q$, first perform polynomial long division.

Types of Partial Fraction Decomposition

Steps for Partial Fractions

• Step 1: Check if the rational function is proper. If not, divide first.
• Step 2: Factorise the denominator $Q(x)$ completely.
• Step 3: Write the appropriate partial fraction form.
• Step 4: Multiply both sides by $Q(x)$ and equate coefficients (or substitute convenient values of $x$) to find constants $A,B,C,\dots$
• Step 5: Integrate each partial fraction separately using standard results.

Worked Example 8 — Distinct Linear Factors

${\displaystyle \int {\displaystyle \frac{2x+1}{(x-1)(x+2)}}dx}$

${\displaystyle \frac{2x+1}{(x-1)(x+2)}}={\displaystyle \frac{A}{x-1}}+{\displaystyle \frac{B}{x+2}}$
Multiply both sides by $(x-1)(x+2)$:
$2x+1=A(x+2)+B(x-1)$
Put $x=1$: $3=3A\Rightarrow A=1$
Put $x=-2$: $-3=-3B\Rightarrow B=1$
${\displaystyle \int {\displaystyle \frac{2x+1}{(x-1)(x+2)}}dx=\int {\displaystyle \frac{dx}{x-1}}+\int {\displaystyle \frac{dx}{x+2}}}$
$=\ln |x-1|+\ln |x+2|+C$
$=\ln |(x-1)(x+2)|+C$.

Worked Example 9 — Repeated Linear Factor

${\displaystyle \int {\displaystyle \frac{x}{(x+1{)}^2}}dx}$

${\displaystyle \frac{x}{(x+1{)}^2}}={\displaystyle \frac{A}{x+1}}+{\displaystyle \frac{B}{(x+1{)}^2}}$
$x=A(x+1)+B$
Put $x=-1$: $-1=B$, so $B=-1$.
Put $x=0$: $0=A+B\Rightarrow A=1$.
${\displaystyle \int {\displaystyle \frac{x}{(x+1{)}^2}}dx=\int {\displaystyle \frac{dx}{x+1}}-\int {\displaystyle \frac{dx}{(x+1{)}^2}}}$
$=\ln |x+1|+{\displaystyle \frac{1}{x+1}}+C$.

Worked Example 10 — Irreducible Quadratic Factor

${\displaystyle \int {\displaystyle \frac{x^2+1}{(x+1)(x^2+1)}}dx}$

Note: $x^2+1$ appears in both numerator and factor, but let's verify:
${\displaystyle \frac{x^2+1}{(x+1)(x^2+1)}}={\displaystyle \frac{1}{x+1}}$   (after cancellation, valid since $x^2+1\ne 0$ for real $x$).
${\displaystyle \int {\displaystyle \frac{1}{x+1}}dx=\ln |x+1|+C}$.