Indefinite Integrals: Standard Forms and Substitution notes: ISC Class 12 Mathematics

1. Integration as anti-differentiation

If $F'(x) = f(x)$ on an interval, $F$ is an antiderivative of $f$ , and $\displaystyle\int f(x)\,dx = F(x) + C$ . Any two antiderivatives differ by a constant, because their difference has derivative $0$ .

$\displaystyle\int [f(x) \pm g(x)]\,dx = \int f(x)\,dx \pm \int g(x)\,dx$
$\displaystyle\int k f(x)\,dx = k\int f(x)\,dx$
A condition such as $F(a) = b$ fixes $C$ .

$y = x^2 - 2$
$y = x^2$
$y = x^2 + 2$

Every member of $x^2 + C$ has derivative $2x$: the curves are vertical shifts of one another.

Imagined figures, in hundred rupees; the constant comes from $P(0) = -50$.
$x$ (hundred cups)	$0$	$1$	$3$	$5$
$P(x) = 40x - 3x^2 - 50$	$-50$	$-13$	$43$	$75$

Formulas

Definition
$\int f(x)\,dx = F(x) + C \iff F'(x) = f(x)$

Key terms

Antiderivative: A function $F$ with $F' = f$ .
Constant of integration: The arbitrary constant $C$ in $F(x) + C$ .

2. Standard integrals and a linear inside

$\displaystyle\int x^n\,dx = \dfrac{x^{n + 1}}{n + 1} + C$ ( $n \ne -1$ ); $\displaystyle\int \dfrac{dx}{x} = \log|x| + C$
$\displaystyle\int e^x\,dx = e^x + C$ ; $\displaystyle\int \sin x\,dx = -\cos x + C$ ; $\displaystyle\int \cos x\,dx = \sin x + C$
$\displaystyle\int \sec^2 x\,dx = \tan x + C$ ; $\displaystyle\int \operatorname{cosec}^2 x\,dx = -\cot x + C$
$\displaystyle\int \sec x\tan x\,dx = \sec x + C$ ; $\displaystyle\int \operatorname{cosec} x\cot x\,dx = -\operatorname{cosec} x + C$

Linear inside: replace $x$ by $ax + b$ and divide by $a$ , e.g. $\displaystyle\int (ax + b)^n\,dx = \dfrac{(ax + b)^{n + 1}}{a(n + 1)} + C$ .

Each row is a derivative read from right to left.
Derivative known	Integral read off
$\dfrac{d}{dx}\,\dfrac{(ax + b)^{n + 1}}{a(n + 1)} = (ax + b)^n$	$\displaystyle\int (ax + b)^n\,dx = \dfrac{(ax + b)^{n + 1}}{a(n + 1)} + C$ , $n \ne -1$
$\dfrac{d}{dx}\log\|x\| = \dfrac1x$	$\displaystyle\int \dfrac{dx}{x} = \log\|x\| + C$
$\dfrac{d}{dx}e^x = e^x$	$\displaystyle\int e^x\,dx = e^x + C$
$\dfrac{d}{dx}(-\cos x) = \sin x$	$\displaystyle\int \sin x\,dx = -\cos x + C$
$\dfrac{d}{dx}\sin x = \cos x$	$\displaystyle\int \cos x\,dx = \sin x + C$
$\dfrac{d}{dx}\tan x = \sec^2 x$	$\displaystyle\int \sec^2 x\,dx = \tan x + C$
$\dfrac{d}{dx}(-\cot x) = \operatorname{cosec}^2 x$	$\displaystyle\int \operatorname{cosec}^2 x\,dx = -\cot x + C$

A linear inside $ax + b$ costs a factor $\dfrac1a$.
Integrand	Integral	Why
$(5x + 2)^3$	$\dfrac{(5x + 2)^4}{20} + C$	divide by $a(n + 1) = 5 \times 4$
$e^{4 - 3x}$	$-\dfrac13 e^{4 - 3x} + C$	divide by the coefficient $-3$
$\cos(2x + 1)$	$\dfrac12\sin(2x + 1) + C$	divide by $2$
$\dfrac{1}{3x - 4}$	$\dfrac13\log\|3x - 4\| + C$	divide by $3$

Checked by differentiating back at four sample points.
Step	Working
Expand	$\dfrac{(x^2 + 1)^2}{x^2} = \dfrac{x^4 + 2x^2 + 1}{x^2} = x^2 + 2 + x^{-2}$
Integrate each term	$\dfrac{x^3}{3} + 2x + \dfrac{x^{-1}}{-1}$
Answer	$\dfrac{x^3}{3} + 2x - \dfrac1x + C$

Formulas

Linear inside
$\int (ax + b)^n\,dx = \frac{(ax + b)^{n + 1}}{a(n + 1)} + C$

Key terms

Standard integral: A result obtained by reading a known derivative backwards.

3. Powers of sine and cosine

Even powers — lower the power: $\sin^2 x = \dfrac{1 - \cos 2x}{2}$ , $\cos^2 x = \dfrac{1 + \cos 2x}{2}$ ; for fourth powers use the identity twice:

$\sin^4 x = \dfrac38 - \dfrac12\cos 2x + \dfrac18\cos 4x$ , $\cos^4 x = \dfrac38 + \dfrac12\cos 2x + \dfrac18\cos 4x$ .

Odd powers — keep one factor: $\sin^3 x = (1 - \cos^2 x)\sin x$ with $t = \cos x$ ; $\cos^3 x = (1 - \sin^2 x)\cos x$ with $t = \sin x$ . The same idea handles products such as $\sin^2 x\cos^3 x$ when one power is odd.

Even powers: lower the power. Odd powers: keep one factor aside.
Power	Rewrite	Integral
$\sin^2 x$	$\dfrac{1 - \cos 2x}{2}$	$\dfrac{x}{2} - \dfrac{\sin 2x}{4} + C$
$\cos^2 x$	$\dfrac{1 + \cos 2x}{2}$	$\dfrac{x}{2} + \dfrac{\sin 2x}{4} + C$
$\sin^3 x$	$(1 - \cos^2 x)\sin x$	$-\cos x + \dfrac{\cos^3 x}{3} + C$
$\cos^3 x$	$(1 - \sin^2 x)\cos x$	$\sin x - \dfrac{\sin^3 x}{3} + C$
$\sin^4 x$	$\dfrac38 - \dfrac12\cos 2x + \dfrac18\cos 4x$	$\dfrac{3x}{8} - \dfrac{\sin 2x}{4} + \dfrac{\sin 4x}{32} + C$
$\cos^4 x$	$\dfrac38 + \dfrac12\cos 2x + \dfrac18\cos 4x$	$\dfrac{3x}{8} + \dfrac{\sin 2x}{4} + \dfrac{\sin 4x}{32} + C$

$y = \sin^2 x$
$y = \dfrac{x}{2} - \dfrac{\sin 2x}{4}$

The antiderivative rises fastest where $\sin^2 x = 1$ and levels off where $\sin^2 x = 0$.

The same steps work for $\sin^3 x$ with $t = \cos x$.

Checked by differentiating back.
Step	Working
Keep one $\sin x$	$\cos^2 x\sin^3 x = \cos^2 x(1 - \cos^2 x)\sin x$
Substitute $t = \cos x$	$dt = -\sin x\,dx$ : $-\displaystyle\int (t^2 - t^4)\,dt$
Integrate	$-\dfrac{t^3}{3} + \dfrac{t^5}{5} + C$
Answer	$-\dfrac{\cos^3 x}{3} + \dfrac{\cos^5 x}{5} + C$

The reduction is an identity: both columns agree.
$x$	$\cos^4 x$	$\tfrac38 + \tfrac12\cos 2x + \tfrac18\cos 4x$
$0.3$	$0.8330$	$0.8330$
$1$	$0.0852$	$0.0852$
$2$	$0.0300$	$0.0300$

Formulas

Square of sine
$\int \sin^2 x\,dx = \frac{x}{2} - \frac{\sin 2x}{4} + C$
Cube of cosine
$\int \cos^3 x\,dx = \sin x - \frac{\sin^3 x}{3} + C$

Key terms

Reduction by identity: Replacing a power of sine or cosine by first powers of cosines of multiple angles.

4. The method of substitution

If $t = g(x)$ then $dt = g'(x)\,dx$ , and $\displaystyle\int f(g(x))g'(x)\,dx = \int f(t)\,dt$ . Choose $t$ so that its derivative is present (up to a constant factor), integrate in $t$ , then return to $x$ .

Useful choices: the inside of a power or a root; the exponent of $e$ ; the argument of a trigonometric function; $\log x$ when $\dfrac1x$ is present; $\sqrt x$ when $\dfrac{1}{\sqrt x}$ is present.

Substitution is the chain rule read backwards.

Which substitution? Ask about derivatives first.

A root of the inside function is a power $n = -\tfrac12$.
Step	Working
Choose	$t = e^{2x} + 3$ , $dt = 2e^{2x}\,dx$
Rewrite	$\displaystyle\int \dfrac{e^{2x}}{\sqrt{e^{2x} + 3}}\,dx = \dfrac12\int t^{-1/2}\,dt$
Integrate	$\dfrac12 \cdot 2t^{1/2} + C = \sqrt{t} + C$
Answer	$\sqrt{e^{2x} + 3} + C$

Formulas

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

Key terms

Substitution: Renaming an inside function $g(x)$ as $t$ , with $dt = g'(x)\,dx$ .

5. The forms f'(x)[f(x)]^n and f'(x)/f(x)

$\displaystyle\int f'(x)[f(x)]^n\,dx = \dfrac{[f(x)]^{n + 1}}{n + 1} + C$ , $n \ne -1$
$\displaystyle\int \dfrac{f'(x)}{f(x)}\,dx = \log|f(x)| + C$

If the numerator is a constant multiple $k\,f'(x)$ , the answer gets the factor $k$ . Drop the modulus only when $f(x) > 0$ throughout (for example $x^2 + 4$ , $e^x + 1$ , $3 + \sin x$ ).

Look for an inside function whose derivative is also present.
Form	Substitution	Result
$f'(x)\,[f(x)]^n$ , $n \ne -1$	$t = f(x)$	$\dfrac{[f(x)]^{n + 1}}{n + 1} + C$
$\dfrac{f'(x)}{f(x)}$	$t = f(x)$	$\log\|f(x)\| + C$
$f'(x)\,e^{f(x)}$	$t = f(x)$	$e^{f(x)} + C$
$f'(x)\cos f(x)$	$t = f(x)$	$\sin f(x) + C$

Checked by differentiating back at three points.
Step	Working
Inside function	$t = x^2 + 3x - 1$
Its derivative	$dt = (2x + 3)\,dx$ — already present
Integrate	$\displaystyle\int t^5\,dt = \dfrac{t^6}{6} + C$
Answer	$\dfrac{(x^2 + 3x - 1)^6}{6} + C$

Adjust a constant factor when the numerator is a multiple of $f'$.
Integrand	Compare with $\dfrac{f'}{f}$	Integral
$\dfrac{x}{x^2 + 4}$	$f = x^2 + 4$ , $f' = 2x$ : half of $\dfrac{f'}{f}$	$\dfrac12\log(x^2 + 4) + C$
$\dfrac{\cos x}{3 + \sin x}$	exactly $\dfrac{f'}{f}$	$\log(3 + \sin x) + C$
$\dfrac{e^{x}}{e^{x} + 1}$	exactly $\dfrac{f'}{f}$	$\log(e^{x} + 1) + C$

Formulas

Logarithmic form
$\int \frac{f'(x)}{f(x)}\,dx = \log|f(x)| + C$

Key terms

f′/f form: A quotient whose numerator is the derivative of its denominator.

6. Integrals of tan x, cot x, sec x and cosec x

$\displaystyle\int \tan x\,dx = -\log|\cos x| + C = \log|\sec x| + C$
$\displaystyle\int \cot x\,dx = \log|\sin x| + C$
$\displaystyle\int \sec x\,dx = \log|\sec x + \tan x| + C$ (multiply by $\dfrac{\sec x + \tan x}{\sec x + \tan x}$ )
$\displaystyle\int \operatorname{cosec} x\,dx = \log|\operatorname{cosec} x - \cot x| + C = \log\left|\tan\dfrac{x}{2}\right| + C$

Each holds on any interval where the integrand is defined.

Each is derived by an $\dfrac{f'}{f}$ substitution.
Integral	Result	Valid where
$\displaystyle\int \tan x\,dx$	$\log\|\sec x\| + C$	$\cos x \ne 0$
$\displaystyle\int \cot x\,dx$	$\log\|\sin x\| + C$	$\sin x \ne 0$
$\displaystyle\int \sec x\,dx$	$\log\|\sec x + \tan x\| + C$	$\cos x \ne 0$
$\displaystyle\int \operatorname{cosec} x\,dx$	$\log\|\operatorname{cosec} x - \cot x\| + C = \log\left\|\tan\dfrac{x}{2}\right\| + C$	$\sin x \ne 0$

$y = \tan x$
$y = \log|\sec x|$

Slope of $\log|\sec x|$ at each $x$ equals $\tan x$.

A clever multiplier creates an $\dfrac{f'}{f}$ form.

Formulas

Secant
$\int \sec x\,dx = \log|\sec x + \tan x| + C$

Key terms

Multiplier method: Multiplying by $\dfrac{u}{u}$ to create an $\dfrac{f'}{f}$ form.

Exam tips

How ISC marks are awarded

Standard integral (1–2 marks): correct form, the factor $\dfrac1a$ for a linear inside, and " $+\,C$ ".
Powers of sine and cosine (2–3 marks): the identity or rewrite, the integration, the simplified answer.
Substitution (2–3 marks): the choice of $t$ and $dt$ , the integral in $t$ , the answer in $x$ .
Derivations of $\int \tan x$ , $\int \sec x$ (3 marks): rewrite or multiplier, substitution, result.
Section D (5 marks): questions come in parts such as 2 + 3; label each part.

Common mistakes

Forgetting " $+\,C$ " or the factor $\dfrac1a$ .
Using the power rule on $\cos^2 x$ as if it were $x^2$ .
Leaving the answer in $t$ .
Dropping the modulus inside $\log$ when the expression can be negative.

1. Integration as anti-differentiation

Formulas

Key terms

2. Standard integrals and a linear inside

Formulas

Key terms

3. Powers of sine and cosine

Formulas

Key terms

4. The method of substitution

Formulas

Key terms

5. The forms f'(x)[f(x)]^n and f'(x)/f(x)

Formulas

Key terms

6. Integrals of tan x, cot x, sec x and cosec x

Formulas

Key terms

Exam tips

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