NDA Maths · Indefinite Integration

Integration by Substitution

Substitution is the reverse chain rule: rename an inner function as u so that its derivative is already sitting in the integrand, collapsing the integral to a standard form in u.

All Indefinite Integration notes NDA Maths strategy

Why this matters

This is the single highest-yield method in the chapter — 17 PYQs, more than any other technique. It splits into a few recognisable shapes: the f′(x)/f(x) → ln pattern, trig-identity reductions, rationalising a surd, and spotting a hidden derivative (xˣ, aˣ). Master the reflex of asking 'is the derivative of some inner part already here?' and most of these become one-liners.

Test yourself — a quick 13-question Indefinite Integration recall quiz, instant score.

Concept 1 of 6: Why Substitution Works — the Reverse Chain Rule

The chain rule says the derivative of a composite carries an extra factor — the derivative of the inside. Substitution undoes exactly that: if the inside's derivative is already in the integrand, renaming the inside as u sweeps it all into a clean integral in u.

Definition

If you set $u = g(x)$ , then $du = g'(x)\,dx$ . The method:

\int f\big(g(x)\big)\,g'(x)\,dx = \int f(u)\,du.

Procedure: (1) choose

u

= the inner function; (2) compute

du = g'(x)\,dx

; (3) replace every

x

-piece so ONLY

u

remains; (4) integrate in

u

; (5) substitute

x

back. If a stray

x

survives step 3, the choice of

u

was wrong. The linear-inner-function shortcut — the $\tfrac1a$ factor three PYQs turn on: if

\int f(x)\,dx=F(x)+C

then

\int f(ax+b)\,dx=\dfrac{1}{a}F(ax+b)+C

, because

u=ax+b

gives

du=a\,dx

. So

\int e^{kx}\,dx=\dfrac{e^{kx}}{k}

\int\cos2x\,dx=\dfrac{\sin2x}{2}

\int\sec^2\tfrac x2\,dx=2\tan\tfrac x2

\int(3x+1)^4\,dx=\dfrac{(3x+1)^5}{15}

\int 5^{2x}\,dx=\dfrac{5^{2x}}{2\ln5}

. Dropping the

\tfrac1a

is the recorded common mistake on one PYQ, and on another the PUBLISHED key is off by exactly this factor of 2 — trust the rule, not the key.

Substitution rule

\int f\big(g(x)\big)\,g'(x)\,dx = \int f(u)\,du,\quad u=g(x)

Worked example

Evaluate

\displaystyle\int 2x\,e^{x^2}\,dx

Practice this concept2 quick reps

Every x must disappear before you integrate in u

After substituting, the integral must be purely in

u

and

du

. A leftover

x

means

u

was chosen badly or a constant factor was mishandled — fix it before integrating.

Concept 2 of 6: Algebraic and Composite Substitutions

When a power of some expression is multiplied by (a constant times) that expression's derivative, substitute the inner expression. This also covers the trick of integrating with respect to a new variable like x-squared.

Definition

Spot a function raised to a power times its derivative — e.g. $\int (\sin x)^3\cos x\,dx$ : set $u=\sin x$ , $du=\cos x\,dx$ , giving $\int u^3\,du$ . The same idea reframes the variable: 'integrate $f$ with respect to $x^2$ ' means $\int f\,d(x^2)$ — treat $t = x^2$ as the variable and use $\int t^n\,dt$ directly.

Power-times-derivative shape

\int \big(g(x)\big)^n\,g'(x)\,dx = \dfrac{\big(g(x)\big)^{n+1}}{n+1} + C

Worked example

Evaluate

\displaystyle\int \cos^4 x\,\sin x\,dx

The same idea in a real exam question:

NDA · Sep 2018 · Q83Easy

Example 2 · Indefinite Integration · Integration by Substitution — Algebraic, Trigonometric, and Composite Forms

What is

\int\sin^3 x\cos x\,dx

equal to?

Carry the sign from du

With

u=\cos x

du=-\sin x\,dx

— the minus sign is part of the substitution. Dropping it flips the answer's sign, and the wrong-sign option is always offered.

Drill 2 more on algebraic and composite substitutions

Concept 3 of 6: The f-prime-over-f to Log Pattern

When the numerator is exactly the derivative of the denominator, the integral is the natural log of the denominator. Train your eye to test the top against the bottom's derivative before doing anything else.

Definition

The single most-tested substitution shape:

\int \dfrac{f'(x)}{f(x)}\,dx = \ln|f(x)| + C.

It is just

u=f(x)

du=f'(x)\,dx

, giving

\int \tfrac{du}{u}

. The disguise is usually in the numerator: differentiate the denominator mentally and check whether the numerator matches (up to a constant you can pull out).

Log pattern

\int \dfrac{f'(x)}{f(x)}\,dx = \ln|f(x)| + C

Worked example

Evaluate

\displaystyle\int \dfrac{2x+1}{x^2+x+5}\,dx

Practice this concept2 quick reps

The same idea in a real exam question:

NDA · Apr 2017 · Q76Moderate

Example 3 · Indefinite Integration · Integration by Substitution — Algebraic, Trigonometric, and Composite Forms

What is

\int\dfrac{(x^{e-1}+e^{x-1})\,dx}{x^e+e^x}

equal to?

Adjust by a constant, never by a variable

If the numerator is

k

times

f'(x)

for a constant

k

, pull

k

out. But if it differs by a FUNCTION of

x

, the pattern does not apply — do not force it.

Drill 1 more on the f-prime-over-f to log pattern

Concept 4 of 6: Trigonometric Substitutions and Identity Reductions

A trig integrand that is not a standard form usually yields to one move: divide through by cos-squared to manufacture sec-squared (the derivative of tan), use a half-angle identity, or simplify a surd of a trig expression into a difference of sine and cosine.

Definition

Three recurring reductions:

Divide by $\cos^2 x$ : for $\int \dfrac{dx}{a^2\sin^2 x + b^2\cos^2 x}$ , divide top and bottom by $\cos^2 x$ to get $\int \dfrac{\sec^2 x\,dx}{a^2\tan^2 x + b^2}$ , then substitute $t=\tan x$ ( $dt=\sec^2x\,dx$ ). The arctan form with a leading coefficient is $\displaystyle\int \dfrac{dt}{a^2t^2+b^2} = \dfrac{1}{ab}\tan^{-1}\!\Big(\dfrac{at}{b}\Big) + C$ — factor $a^2$ out first: $\dfrac{1}{a^2}\int\dfrac{dt}{t^2+(b/a)^2} = \dfrac{1}{a^2}\cdot\dfrac{a}{b}\tan^{-1}\dfrac{at}{b}$ . So the whole integral is $\dfrac{1}{ab}\tan^{-1}\!\Big(\dfrac{a\tan x}{b}\Big) + C$ .
Half-angle: $1-\cos x = 2\sin^2\tfrac{x}{2}$ and $1+\cos x = 2\cos^2\tfrac{x}{2}$ turn a fraction into a $\csc^2$ or $\sec^2$ you can integrate.
Surd identities: $1 \pm \sin 2x = (\sin x \pm \cos x)^2$ , so $\sqrt{1-\sin 2x} = |\sin x - \cos x|$ becomes integrable. The simplification $\sec x + \tan x$ inside an inverse-tan also collapses to $\tan(\tfrac{\pi}{4}+\tfrac{x}{2})$ .

The divide-by-cos-squared move

\int \dfrac{dx}{a^2\sin^2 x + b^2\cos^2 x} = \int \dfrac{\sec^2 x\,dx}{a^2\tan^2 x + b^2} = \dfrac{1}{ab}\tan^{-1}\!\Big(\dfrac{a\tan x}{b}\Big) + C

Worked example

Evaluate

\displaystyle\int \dfrac{dx}{1+3\cos^2 x}

The same idea in a real exam question:

NDA · Sep 2018 · Q97Moderate

Example 4 · Indefinite Integration · Integration by Substitution — Algebraic, Trigonometric, and Composite Forms

What is

\int\frac{dx}{a^2\sin^2x+b^2\cos^2x}

equal to?

Which of $a$ , $b$ goes inside the arctan?

\int \dfrac{dt}{a^2t^2+b^2}

the coefficient of

t^2

ends up multiplying

t

inside:

\tan^{-1}\big(\tfrac{at}{b}\big)

, not

\tan^{-1}\big(\tfrac{bt}{a}\big)

. Check with

a=1

\int\frac{dt}{t^2+b^2} = \frac1b\tan^{-1}\frac{t}{b}

—

t

is divided by the constant term's root. A published key for this exact PYQ had

a

and

b

swapped; the

\tfrac{1}{ab}

outside is the same either way, so only the inside tells the options apart.

A square root forces an absolute value

\sqrt{(\sin x - \cos x)^2} = |\sin x - \cos x|

, and the sign depends on the given range of

x

. On

0<x<\tfrac{\pi}{4}

\cos x > \sin x

, so it equals

\cos x - \sin x

— read the interval before dropping the modulus.

Drill 7 more on trigonometric substitutions and identity reductions

Concept 5 of 6: Rationalising a Surd Denominator

A difference of two square roots in the denominator clears the moment you multiply by its conjugate — the difference of squares wipes out the surds and leaves a simple sum of powers to integrate.

Definition

For $\displaystyle\int \dfrac{dx}{\sqrt{x+a}-\sqrt{x+b}}$ , multiply numerator and denominator by the conjugate $\sqrt{x+a}+\sqrt{x+b}$ . The denominator becomes $(x+a)-(x+b) = a-b$ , a constant, leaving $\dfrac{1}{a-b}\int\big(\sqrt{x+a}+\sqrt{x+b}\big)\,dx$ — two power-rule integrals. Each $\int (x+c)^{1/2}\,dx = \tfrac{2}{3}(x+c)^{3/2}$ .

Conjugate clears the surd

\dfrac{1}{\sqrt{x+a}-\sqrt{x+b}} = \dfrac{\sqrt{x+a}+\sqrt{x+b}}{a-b}

Worked example

Evaluate

\displaystyle\int \dfrac{dx}{\sqrt{x+3}-\sqrt{x}}

The same idea in a real exam question:

NDA · Apr 2024 · Q95Moderate

Example 5 · Indefinite Integration · Integration by Substitution — Algebraic, Trigonometric, and Composite Forms

Let

\displaystyle\int\dfrac{dx}{\sqrt{x+1}-\sqrt{x-1}}=\alpha(x+1)^{3/2}+\beta(x-1)^{3/2}+c

What is the value of

\alpha

Do not lose the 1 over (a minus b) factor

After conjugating, the constant denominator

a-b

stays as a multiplier on the whole integral. Forgetting it scales every coefficient wrong — the classic error on

\sqrt{x+1}-\sqrt{x-1}

(where

a-b=2

Drill 1 more on rationalising a surd denominator

Concept 6 of 6: Spotting a Hidden Derivative

The hardest substitutions hide the derivative of an unusual function inside the integrand — the derivative of x-to-the-x, or an exponential base whose derivative is itself times a constant. Recognise the derivative, name its parent as u, and the integral collapses.

Definition

Useful hidden derivatives:

$\dfrac{d}{dx}\,x^x = x^x(1+\ln x)$ . So $\int (x^x)^2(1+\ln x)\,dx$ : set $u=x^x$ , $du = x^x(1+\ln x)\,dx$ , giving $\int u\,du = \tfrac{u^2}{2}$ .
$\dfrac{d}{dx}\,a^x = a^x\ln a$ . For $\int \dfrac{dx}{a^x - 1}$ or $\int \dfrac{dx}{a^x + a^{-x}}$ , multiply through by $a^x$ and set $u=a^x$ ( $du = a^x\ln a\,dx$ ) to reach a rational or arctan form in $u$ .

The x-to-the-x derivative

\dfrac{d}{dx}\,x^x = x^x(1+\ln x)

Worked example

Evaluate

\displaystyle\int \dfrac{4^x}{4^{x}+1}\,dx

The same idea in a real exam question:

NDA · Sep 2022 · Q74Hard

Example 6 · Indefinite Integration · Integration by Substitution — Algebraic, Trigonometric, and Composite Forms

What is

\int (x^x)^2(1+\ln x)\,dx

equal to?

x-to-the-x is neither a power nor an exponential

\dfrac{d}{dx}x^x \neq x\cdot x^{x-1}

(power rule) and

\neq x^x\ln x

(exponential rule). Both the base and the exponent vary, so its derivative is

x^x(1+\ln x)

— derived via logarithmic differentiation.

Drill 2 more on spotting a hidden derivative

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (6)

Why Substitution Works — the Reverse Chain Rule
Substitution rule
$\int f\big(g(x)\big)\,g'(x)\,dx = \int f(u)\,du,\quad u=g(x)$
Algebraic and Composite Substitutions
Power-times-derivative shape
$\int \big(g(x)\big)^n\,g'(x)\,dx = \dfrac{\big(g(x)\big)^{n+1}}{n+1} + C$
The f-prime-over-f to Log Pattern
Log pattern
$\int \dfrac{f'(x)}{f(x)}\,dx = \ln|f(x)| + C$
Trigonometric Substitutions and Identity Reductions
The divide-by-cos-squared move
$\int \dfrac{dx}{a^2\sin^2 x + b^2\cos^2 x} = \int \dfrac{\sec^2 x\,dx}{a^2\tan^2 x + b^2} = \dfrac{1}{ab}\tan^{-1}\!\Big(\dfrac{a\tan x}{b}\Big) + C$
Rationalising a Surd Denominator
Conjugate clears the surd
$\dfrac{1}{\sqrt{x+a}-\sqrt{x+b}} = \dfrac{\sqrt{x+a}+\sqrt{x+b}}{a-b}$
Spotting a Hidden Derivative
The x-to-the-x derivative
$\dfrac{d}{dx}\,x^x = x^x(1+\ln x)$

Watch out for (7)

Every x must disappear before you integrate in u
→ Why Substitution Works — the Reverse Chain Rule
Carry the sign from du
→ Algebraic and Composite Substitutions
Adjust by a constant, never by a variable
→ The f-prime-over-f to Log Pattern
Which of $a$ , $b$ goes inside the arctan?
→ Trigonometric Substitutions and Identity Reductions
A square root forces an absolute value
→ Trigonometric Substitutions and Identity Reductions
Do not lose the 1 over (a minus b) factor
→ Rationalising a Surd Denominator
x-to-the-x is neither a power nor an exponential
→ Spotting a Hidden Derivative

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