Integration by Substitution

The Substitution Method

Inspection runs out of steam quickly — we need a systematic technique. Substitution is the chain rule played backwards: transform $\int f(x)\,dx$ by changing the variable from $x$ to $t$ via $x = g(t)$, so that $dx = g'(t)\,dt$ and

$$\int f(x)\,dx = \int f(g(t))\,g'(t)\,dt$$

In practice you usually spot it the other way round: substitute $t = (\text{some inner function})$ whose derivative also sits in the integrand.

The 4-step recipe

1. Spot the inner function whose derivative (up to a constant) appears as a factor: in $\int 2x\sin(x^2+1)\,dx$, the inner function is $x^2 + 1$ and its derivative $2x$ is right there.
2. Substitute: $t = x^2 + 1$, $dt = 2x\,dx$ — the integral collapses to $\int \sin t\,dt$.
3. Integrate in $t$: $-\cos t + C$.
4. Return to $x$: $-\cos(x^2 + 1) + C$. Always convert back — an answer left in $t$ is incomplete.

If the derivative appears with a wrong constant, adjust: for $\int \sin mx\,dx$, put $t = mx$, $dt = m\,dx$, so $dx = \frac{dt}{m}$ and the answer is $-\frac{1}{m}\cos mx + C$.

Key Point: Substitution works when the integrand is (inner function's derivative) × (function of the inner function). Train your eye to scan for the pair $g(x)$ and $g'(x)$ before anything else.

[JEE Tip] Common inner functions with instantly recognisable derivatives: $x^2 + a$ (deriv $2x$), $\log x$ (deriv $\frac{1}{x}$), $\tan^{-1}x$ (deriv $\frac{1}{1+x^2}$), $\sin^{-1}x$ (deriv $\frac{1}{\sqrt{1-x^2}}$), $\sqrt{x}$ (deriv $\frac{1}{2\sqrt x}$), $e^x$, $\sin x/\cos x$ pairs. Half of JEE substitution questions use one of these seven.

The Four Famous Results

Substitution immediately delivers four integrals used constantly for the rest of the chapter — memorise them as table entries:

1. $\displaystyle\int \tan x\,dx = \log|\sec x| + C$

Proof: $\int \frac{\sin x}{\cos x}dx$; put $t = \cos x$, $dt = -\sin x\,dx$: $-\int\frac{dt}{t} = -\log|\cos x| + C = \log|\sec x| + C$.

2. $\displaystyle\int \cot x\,dx = \log|\sin x| + C$

Proof: $\int\frac{\cos x}{\sin x}dx$; put $t = \sin x$, $dt = \cos x\,dx$: $\int\frac{dt}{t} = \log|\sin x| + C$.

3. $\displaystyle\int \sec x\,dx = \log|\sec x + \tan x| + C$

Proof (the famous trick): multiply and divide by $\sec x + \tan x$: $$\int \frac{\sec x(\sec x + \tan x)}{\sec x + \tan x}dx$$ Put $t = \sec x + \tan x$; then $dt = (\sec x\tan x + \sec^2 x)\,dx$ — exactly the numerator! So the integral is $\int\frac{dt}{t} = \log|\sec x + \tan x| + C$.

4. $\displaystyle\int \text{cosec}\,x\,dx = \log|\text{cosec}\,x - \cot x| + C$

Proof: same trick with $\text{cosec}\,x + \cot x$, then simplify using $\text{cosec}^2x - \cot^2 x = 1$.

Key Point: The $\sec$ and $\text{cosec}$ derivations are exam questions in their own right ("Prove that $\int \sec x\,dx = \ldots$"). Learn the multiply-and-divide trick, not just the results.

[Board Important] Write the modulus signs in all four results. $\log \sec x$ without modulus loses a mark whenever $\sec x$ can be negative.

Harder Substitution Patterns

Pattern 1 — Odd powers of sine or cosine: peel one factor off.

For $\int \sin^3 x\cos^2 x\,dx$: the power of sine is odd, so write $\sin^3 x = (1 - \cos^2 x)\sin x$ and substitute $t = \cos x$:

$$\int (1 - t^2)t^2(-dt) = -\frac{t^3}{3}\cdot(-1)\ldots = -\frac{\cos^3 x}{3} + \frac{\cos^5 x}{5} + C$$

Rule of thumb: odd power of $\sin$ → substitute $t = \cos x$; odd power of $\cos$ → substitute $t = \sin x$; both even → use identities (next block).

Pattern 2 — Shifted angles: substitute the shift.

For $\int \frac{\sin x}{\sin(x + a)}dx$: put $t = x + a$, expand $\sin(t - a) = \sin t\cos a - \cos t\sin a$:

$$\int \frac{\sin(t-a)}{\sin t}dt = \cos a\int dt - \sin a\int \cot t\,dt = x\cos a - \sin a\log|\sin(x+a)| + C$$

Pattern 3 — $\frac{1}{1 + \tan x}$ types: split the numerator smartly.

$\int \frac{dx}{1 + \tan x} = \int\frac{\cos x\,dx}{\cos x + \sin x}$. Write the numerator as a combination of the denominator and its derivative:

$$\cos x = \frac{1}{2}(\cos x + \sin x) + \frac{1}{2}(\cos x - \sin x)$$

The first piece integrates to $\frac{x}{2}$; the second is $\frac{1}{2}\int\frac{d(\cos x + \sin x)}{\cos x + \sin x} = \frac{1}{2}\log|\cos x + \sin x|$. Total: $\frac{x}{2} + \frac{1}{2}\log|\cos x + \sin x| + C$.

Key Point: Pattern 3's idea — numerator = A(denominator) + B(derivative of denominator) — is a general JEE technique for $\int\frac{a\sin x + b\cos x}{c\sin x + d\cos x}dx$. Solve for A and B by comparing coefficients.

[JEE Tip] In Pattern 1, if BOTH powers are odd, peel from the smaller power — less algebra. If the integrand is $\sin^m x\cos^n x$ with $m + n$ even and negative-ready, JEE-level shortcuts exist via $t = \tan x$ — see the JEE Corner.

Integration Using Trigonometric Identities

When the integrand is built from powers or products of sines and cosines with NO convenient derivative pair, reshape it with identities until every term is linear in some $\sin(kx)$ or $\cos(kx)$.

Power reduction (even powers):

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

So $\int \cos^2 x\,dx = \frac{x}{2} + \frac{\sin 2x}{4} + C$, and fourth powers just need the identity twice: $\sin^4 x = \left(\frac{1-\cos 2x}{2}\right)^2$, then reduce $\cos^2 2x$ again.

Cubes (odd powers) via triple-angle:

$$\sin^3 x = \frac{3\sin x - \sin 3x}{4}, \qquad \cos^3 x = \frac{3\cos x + \cos 3x}{4}$$

giving $\int \sin^3 x\,dx = -\frac{3}{4}\cos x + \frac{1}{12}\cos 3x + C$ (equivalently $-\cos x + \frac{\cos^3 x}{3} + C$ by the peel-a-factor route — both are correct, differing by a constant… in fact by identity they are equal!).

Products into sums:

$$\sin x\cos y = \tfrac{1}{2}[\sin(x+y) + \sin(x-y)], \quad \cos x\cos y = \tfrac{1}{2}[\cos(x+y) + \cos(x-y)], \quad \sin x\sin y = \tfrac{1}{2}[\cos(x-y) - \cos(x+y)]$$

So $\int \sin 2x\cos 3x\,dx = \frac{1}{2}\int[\sin 5x - \sin x]\,dx = -\frac{\cos 5x}{10} + \frac{\cos x}{2} + C$.

Half-angle for $1 \pm \cos x$:

$$1 + \cos x = 2\cos^2\tfrac{x}{2}, \qquad 1 - \cos x = 2\sin^2\tfrac{x}{2}$$

These demolish integrands like $\frac{1-\cos x}{1+\cos x} = \tan^2\frac{x}{2}$.

Key Point: The goal of every identity manoeuvre is the same: reach a sum of terms of the form $a\sin(kx)$, $b\cos(kx)$ or constants — each integrable in one line with the divide-by-$k$ reflex.

[Board Important] For $\int \sin A\cos B$ products, expect half-marks reserved for correctly writing the product-to-sum identity. Quote it, substitute, then integrate — don't jump.

Solved Examples

Example 1: The basic adjust-the-constant substitution

Integrate $\sin mx$ with respect to $x$.

Solution:

1. Substitute: $t = mx$, so $dt = m\,dx$, i.e. $dx = \frac{dt}{m}$.
2. Transform: $\int \sin mx\,dx = \frac{1}{m}\int \sin t\,dt = -\frac{1}{m}\cos t + C$.
3. Return to $x$: $-\frac{1}{m}\cos mx + C$.

Final Answer: $-\frac{1}{m}\cos mx + C$.

Takeaway: Linear inner functions just cost a division by the coefficient — soon you will do this without writing the substitution at all.

Example 2: The model derivative-pair substitution

Find $\int 2x \sin(x^2 + 1)\,dx$.

Solution:

1. Spot the pair: inner function $x^2 + 1$ has derivative $2x$ — present as a factor.
2. Substitute: $t = x^2 + 1$, $dt = 2x\,dx$: the integral becomes $\int \sin t\,dt$.
3. Integrate and return: $-\cos t + C = -\cos(x^2 + 1) + C$.

Final Answer: $-\cos(x^2 + 1) + C$.

Takeaway: This is the picture-perfect substitution: derivative sitting beside its function. Most exam integrands are disguised versions of exactly this.

Example 3: Substitution inside substitution

Find $\int \frac{\tan^4\sqrt{x}\,\sec^2\sqrt{x}}{\sqrt{x}}\,dx$.

Solution:

1. First substitution: $t = \sqrt x$, $dt = \frac{dx}{2\sqrt x}$, so $\frac{dx}{\sqrt x} = 2\,dt$: the integral is $2\int \tan^4 t\sec^2 t\,dt$.
2. Second substitution: $u = \tan t$, $du = \sec^2 t\,dt$: it becomes $2\int u^4 du = \frac{2u^5}{5}$.
3. Unwind both: $\frac{2}{5}\tan^5 t = \frac{2}{5}\tan^5\sqrt{x} + C$.

Final Answer: $\frac{2}{5}\tan^5\sqrt{x} + C$.

Takeaway: Layered integrands unwrap one layer at a time — and the single substitution $u = \tan\sqrt x$ would have done it in one step. Practice builds that compression.

Example 4: An inverse-trig inner function

Find $\int \frac{\sin(\tan^{-1}x)}{1 + x^2}\,dx$.

Solution:

1. Spot the pair: $\tan^{-1}x$ has derivative $\frac{1}{1+x^2}$ — present.
2. Substitute: $t = \tan^{-1}x$, $dt = \frac{dx}{1+x^2}$: the integral is $\int \sin t\,dt = -\cos t + C$.
3. Return: $-\cos(\tan^{-1}x) + C$.

Final Answer: $-\cos(\tan^{-1}x) + C$.

Takeaway: $\frac{1}{1+x^2}$ in an integrand is a flashing sign saying "substitute $t = \tan^{-1}x$".

Example 5: Deriving the famous $\int \tan x$ and $\int \sec x$

Prove that (i) $\int \tan x\,dx = \log|\sec x| + C$ and (ii) $\int \sec x\,dx = \log|\sec x + \tan x| + C$.

Solution:

1. (i) Write as a fraction: $\int \frac{\sin x}{\cos x}dx$. Put $t = \cos x$, $dt = -\sin x\,dx$.
2. Integrate: $-\int\frac{dt}{t} = -\log|t| + C = -\log|\cos x| + C = \log|\sec x| + C$.
3. (ii) The multiply-divide trick: $\int \sec x\,dx = \int\frac{\sec x(\sec x + \tan x)}{\sec x + \tan x}dx$.
4. Substitute the denominator: $t = \sec x + \tan x$ gives $dt = (\sec x\tan x + \sec^2 x)dx$ — precisely the numerator.
5. Integrate: $\int\frac{dt}{t} = \log|t| + C = \log|\sec x + \tan x| + C$. ∎

Takeaway: (ii) is a Board favourite "prove that" question. The whole trick is manufacturing the derivative of the denominator in the numerator.

Example 6: Odd power — peel a factor

Find $\int \sin^3 x\cos^2 x\,dx$.

Solution:

1. Peel one sine: $\sin^3 x\cos^2 x = (1 - \cos^2 x)\cos^2 x\,\sin x$.
2. Substitute: $t = \cos x$, $dt = -\sin x\,dx$: the integral is $-\int(1 - t^2)t^2\,dt = -\int(t^2 - t^4)dt$.
3. Integrate: $-\left(\frac{t^3}{3} - \frac{t^5}{5}\right) + C$.
4. Return: $-\frac{\cos^3 x}{3} + \frac{\cos^5 x}{5} + C$.

Final Answer: $-\frac{1}{3}\cos^3 x + \frac{1}{5}\cos^5 x + C$.

Takeaway: Odd power of $\sin$ → save one $\sin x$ for $dt$, convert the rest to cosines. The parity decides the substitution for you.

Example 7: The shifted-angle pattern

Find $\int \frac{\sin x}{\sin(x + a)}\,dx$.

Solution:

1. Substitute the shift: $t = x + a$, $dx = dt$: the integral is $\int\frac{\sin(t - a)}{\sin t}dt$.
2. Expand the numerator: $\sin(t-a) = \sin t\cos a - \cos t\sin a$, so the integrand splits as $\cos a - \sin a\cot t$.
3. Integrate: $t\cos a - \sin a\log|\sin t| + C_1$.
4. Return and absorb constants: $(x+a)\cos a - \sin a\log|\sin(x+a)| + C_1 = x\cos a - \sin a\log|\sin(x+a)| + C$.

Final Answer: $x\cos a - \sin a\,\log|\sin(x + a)| + C$.

Takeaway: The constant $a\cos a$ melted into $C$ — arbitrary constants absorb every fixed number. Expect the same in the $\frac{\cos x}{\cos(x-a)}$ family.

Example 8: Numerator = A(denominator) + B(derivative)

Find $\int \frac{dx}{1 + \tan x}$.

Solution:

1. Clear the tan: $\frac{1}{1+\tan x} = \frac{\cos x}{\cos x + \sin x}$.
2. Split the numerator: $\cos x = \frac{1}{2}(\cos x + \sin x) + \frac{1}{2}(\cos x - \sin x)$ — half the denominator plus half its derivative.
3. Integrate the first piece: $\frac{1}{2}\int dx = \frac{x}{2}$.
4. Integrate the second piece: $\frac{1}{2}\int\frac{\cos x - \sin x}{\cos x + \sin x}dx$; with $t = \cos x + \sin x$, $dt = (\cos x - \sin x)dx$: $= \frac{1}{2}\log|\cos x + \sin x|$.

Final Answer: $\frac{x}{2} + \frac{1}{2}\log|\cos x + \sin x| + C$.

Takeaway: For $\int\frac{a\sin x + b\cos x}{c\sin x + d\cos x}dx$, always try numerator $= A(\text{denominator}) + B(\text{denominator})'$ — a two-unknown comparison that JEE reuses constantly.

Example 9: Even power — reduce it

Find $\int \cos^2 x\,dx$.

Solution:

1. Power-reduce: $\cos^2 x = \frac{1 + \cos 2x}{2}$.
2. Integrate: $\frac{1}{2}\int dx + \frac{1}{2}\int \cos 2x\,dx = \frac{x}{2} + \frac{1}{2}\cdot\frac{\sin 2x}{2}$.

Final Answer: $\frac{x}{2} + \frac{\sin 2x}{4} + C$.

Takeaway: Even powers cannot be peeled — they must be reduced. $\frac{x}{2} + \frac{\sin 2x}{4}$ is worth memorising; it appears inside dozens of longer problems.

Example 10: Product to sum

Find $\int \sin 2x\cos 3x\,dx$.

Solution:

1. Apply the identity $\sin A\cos B = \frac{1}{2}[\sin(A+B) + \sin(A-B)]$: $\sin 2x\cos 3x = \frac{1}{2}[\sin 5x + \sin(-x)] = \frac{1}{2}(\sin 5x - \sin x)$.
2. Integrate term by term: $\frac{1}{2}\left[-\frac{\cos 5x}{5} + \cos x\right] + C$.

Final Answer: $-\frac{1}{10}\cos 5x + \frac{1}{2}\cos x + C$.

Takeaway: $\sin(-x) = -\sin x$ — the sign inside the product-to-sum identity is where most marks are lost.

Example 11: $\sin^3 x$ two ways

Find $\int \sin^3 x\,dx$ (i) using the triple-angle identity, (ii) by peeling a factor.

Solution:

1. (i) Triple angle: from $\sin 3x = 3\sin x - 4\sin^3 x$: $\sin^3 x = \frac{3\sin x - \sin 3x}{4}$.
2. Integrate: $\frac{3}{4}\int\sin x\,dx - \frac{1}{4}\int\sin 3x\,dx = -\frac{3}{4}\cos x + \frac{1}{12}\cos 3x + C$.
3. (ii) Peel: $\sin^3 x = (1 - \cos^2 x)\sin x$; with $t = \cos x$: $-\int(1 - t^2)dt = -t + \frac{t^3}{3}$, i.e. $-\cos x + \frac{\cos^3 x}{3} + C$.
4. Reconcile: expanding $\cos 3x = 4\cos^3 x - 3\cos x$ in (i) reproduces (ii) exactly — the two answers are identical, not merely equivalent.

Final Answer: $-\frac{3}{4}\cos x + \frac{1}{12}\cos 3x + C$ $\;\left(= -\cos x + \frac{\cos^3 x}{3} + C\right)$.

Takeaway: Different legitimate methods can produce different-looking answers. Differentiate (or use identities) before suspecting an error.

Example 12: Log and inverse-tan pairs

Find (i) $\int \frac{(\log x)^2}{x}\,dx$ (ii) $\int \frac{e^{\tan^{-1}x}}{1 + x^2}\,dx$.

Solution:

1. (i) Substitute: $t = \log x$, $dt = \frac{dx}{x}$: $\int t^2 dt = \frac{t^3}{3} + C = \frac{(\log x)^3}{3} + C$.
2. (ii) Substitute: $t = \tan^{-1}x$, $dt = \frac{dx}{1+x^2}$: $\int e^t dt = e^t + C = e^{\tan^{-1}x} + C$.

Final Answer: (i) $\frac{(\log x)^3}{3} + C$; (ii) $e^{\tan^{-1}x} + C$.

Takeaway: $\frac{1}{x}$ next to a function of $\log x$, and $\frac{1}{1+x^2}$ next to a function of $\tan^{-1}x$ — two of the most reused derivative pairs in exams.

Example 13: A fourth power, fully reduced

Find $\int \sin^4 x\,dx$.

Solution:

1. Reduce once: $\sin^4 x = \left(\frac{1 - \cos 2x}{2}\right)^2 = \frac{1 - 2\cos 2x + \cos^2 2x}{4}$.
2. Reduce the leftover square: $\cos^2 2x = \frac{1 + \cos 4x}{2}$, so $\sin^4 x = \frac{1}{4} - \frac{\cos 2x}{2} + \frac{1}{8} + \frac{\cos 4x}{8} = \frac{3}{8} - \frac{\cos 2x}{2} + \frac{\cos 4x}{8}$.
3. Integrate term by term: $\frac{3x}{8} - \frac{\sin 2x}{4} + \frac{\sin 4x}{32} + C$.

Final Answer: $\frac{3x}{8} - \frac{\sin 2x}{4} + \frac{\sin 4x}{32} + C$.

Takeaway: Each application of power reduction halves the power and doubles the angle. Two rounds turn any fourth power into one-line integrals.