Integration by Substitution

MATStandard Forms & Substitution

Integration by substitution reverses the chain rule: you replace a chunk of the integrand by a new variable $t$ so that its differential $dt$ is also present, turning a hard integral into a standard one. The signal to use it is that the integrand contains a function and (a multiple of) its derivative — for example $\cos x\,e^{\sin x}$ or $\tan^3 x\sec^2 x$ .

ISC tests it constantly because almost every other technique (parts, partial fractions, standard forms) ultimately leans on a clean substitution.

Substitution rule

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

t=g(x)

dt = g'(x)\,dx

; choose

g

as the inner function whose derivative already appears (up to a constant).

Logarithmic form (derivative on top)

\int \frac{g'(x)}{g(x)}\,dx = \ln|g(x)| + C

When the numerator is exactly the derivative of the denominator; e.g.

\int\dfrac{2x+3}{x^2+3x+2}\,dx=\ln|x^2+3x+2|+C

Power of a function

\int [g(x)]^{n}\,g'(x)\,dx = \frac{[g(x)]^{n+1}}{n+1} + C, \quad n \ne -1

E.g.

\int \tan^3 x\sec^2 x\,dx

with

t=\tan x

dt=\sec^2 x\,dx

gives

\dfrac{\tan^4 x}{4}+C

Exponential composite

\int e^{g(x)}\,g'(x)\,dx = e^{g(x)} + C

E.g.

\int \cos x\,e^{\sin x}\,dx = e^{\sin x}+C

with

t=\sin x

Definite integral: change the limits

\int_{a}^{b} f(g(x))\,g'(x)\,dx = \int_{g(a)}^{g(b)} f(t)\,dt

When

t=g(x)

, replace limits

x=a,b

t=g(a),g(b)

; then no back-substitution is needed.

Choose the substitution so that $dt=g'(x)\,dx$ is already (a constant multiple of) a factor in the integrand; adjust the constant outside the integral.
After substituting, the integral must contain only the new variable $t$ — if any $x$ remains, the substitution is incomplete or wrong.
For $\int\dfrac{\sin x}{1+\cos x}\,dx$ take $t=1+\cos x$ , $dt=-\sin x\,dx$ , giving $-\ln|1+\cos x|+C$ .
Standard trig substitutions: $\sqrt{a^2-x^2}\Rightarrow x=a\sin\theta$ ; $\sqrt{a^2+x^2}\Rightarrow x=a\tan\theta$ ; $\sqrt{x^2-a^2}\Rightarrow x=a\sec\theta$ .
Rationalising substitution for $\dfrac{1}{1+\sqrt{x}}$ : put $x=t^2$ ( $t\ge 0$ ), $dx=2t\,dt$ , turning the surd into a rational function.
For indefinite integrals, always substitute back to the original variable and add $+C$ ; for definite integrals, change the limits instead and skip back-substitution.
Recognise $\int \dfrac{g'(x)}{g(x)}\,dx=\ln|g(x)|+C$ on sight — top is the derivative of the bottom — to avoid an unnecessary formal substitution.

Where the marks go

Forgetting to transform $dx$ into $dt$ (dropping the $g'(x)$ factor), which makes the answer wrong.
In a definite integral, changing the variable but keeping the old $x$ -limits — the limits must become $t$ -limits $g(a),g(b)$ .
Leaving the answer in terms of $t$ for an indefinite integral instead of returning to $x$ , or omitting $+C$ .
Missing the absolute value in $\ln|g(x)|$ , which is required wherever $g(x)$ can be negative.

How the board asks it

Numericalreverse chain rule; integrand holds $g(x)$ and a multiple of $g'(x)$
Find $\int \cos x\, e^{\sin x}\,dx$ .
Numerical $\int \dfrac{g'(x)}{g(x)}\,dx=\ln|g(x)|+C$ recognised on sight
Evaluate $\int \dfrac{2x+1}{x^2+x+1}\,dx$ .
Numericaldefinite integral; transform limits with the variable, skip back-substitution
Evaluate $\int_{0}^{\pi/2} \dfrac{\sin x}{1+\cos^2 x}\,dx$ by changing the variable and transforming the limits accordingly.
Numericalstandard substitutions: $\sqrt{a^2-x^2}\Rightarrow x=a\sin\theta$
Using a suitable substitution, evaluate $\int \dfrac{dx}{\sqrt{a^2-x^2}}$ by putting $x=a\sin\theta$ .
Numericalput $x=t^2$ to turn $\sqrt{x}$ into a rational function
Evaluate $\int \dfrac{1}{1+\sqrt{x}}\,dx$ using the substitution $x=t^2$ .

Practise this topic

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Written for Sublevo. Question text quoted anywhere in these notes is the Council’s and carries its year and paper; the board’s own diagrams are not reproduced.