Integration by substitutionEdexcel International A Level Maths: Revision notes

Section 1

The idea: the reverse of the chain rule

Integration by substitution reverses the chain rule. If an integral contains a function of a function multiplied by (a multiple of) the derivative of the inner function, replace the inner function by $u$ : $\int f\big(g(x)\big)\,g'(x)\,dx=\int f(u)\,du,\qquad u=g(x),\; du=g'(x)\,dx.$ Example: $I=\int 2x(x^2+3)^4\,dx$ . Let $u=x^2+3$ , so $\frac{du}{dx}=2x$ and $du=2x\,dx$ . Then $I=\int u^4\,du=\frac{u^5}{5}+c=\frac15(x^2+3)^5+c$ . Always write the final answer back in terms of $x$ for an indefinite integral.

Key termssubstitutionreverse chain rule

Common mistake

Forgetting to replace $dx$ . The $dx$ must become an expression in $du$ so that nothing in $x$ is left.

Section 2

Dealing with a leftover x

Sometimes $x$ is still present after substituting. Make $x$ the subject of the substitution and replace it too. Example: $\int x\sqrt{x-2}\,dx$ with $u=x-2$ . Then $x=u+2$ and $dx=du$ : $\int(u+2)\sqrt u\,du=\int\left(u^{3/2}+2u^{1/2}\right)du=\frac25u^{5/2}+\frac43u^{3/2}+c.$ Finally substitute $u=x-2$ : $\frac25(x-2)^{5/2}+\frac43(x-2)^{3/2}+c$ . Expand brackets before integrating so that each term is a power of $u$ .

Key termsleftover x

Exam tip

If the substitution is $u=ax+b$ then $dx=\frac{du}{a}$ . Do not forget the $\frac1a$ .

Section 3

Definite integrals: change the limits

For a definite integral you can change the limits to match the new variable, so you never have to go back to $x$ . If $u=g(x)$ then the limits $x=a$ and $x=b$ become $u=g(a)$ and $u=g(b)$ . Example: $\int_0^12x(x^2+3)^4\,dx$ with $u=x^2+3$ . At $x=0$ , $u=3$ ; at $x=1$ , $u=4$ . $\int_3^4u^4\,du=\left[\frac{u^5}{5}\right]_3^4=\frac{1024-243}{5}=\frac{781}{5}.$ Alternatively integrate, return to $x$ and use the original limits. Do not mix the two: $u$ -limits with an $x$ answer is a common slip.

Key termslimits

Common mistake

Using the original $x$ -limits with an expression in $u$ .

Section 4

When the substitution is given

In harder integrals the question states the substitution. Follow it exactly: differentiate to find $du$ in terms of $dx$ (or $dx$ in terms of $du$ ), rewrite $x$ and the rest of the integrand, change the limits, then integrate. Example: $I=\int_0^4\frac{x}{\sqrt{2x+1}}\,dx$ with $u=2x+1$ . Then $x=\frac{u-1}{2}$ , $dx=\frac12du$ , and the limits are $u=1$ and $u=9$ : $I=\frac14\int_1^9\left(u^{1/2}-u^{-1/2}\right)du=\frac14\left[\frac23u^{3/2}-2u^{1/2}\right]_1^9=\frac{10}{3}.$ Area problems: if $u=\sqrt{x+1}$ then $x=u^2-1$ and $dx=2u\,du$ , which often cancels a denominator.

Exam tip

Check an answer by differentiating it: you should recover the original integrand.

Section 5

Exam technique

Show the substitution line by line: $u=\dots$ , $\frac{du}{dx}=\dots$ , new limits, new integrand. In a 'show that' question the result is given, so every step must appear. Leave exact answers as fractions or surds. Check that no $x$ remains inside an integral with respect to $u$ .

Common mistake

Substituting for $x$ in the integrand but leaving the $dx$ unchanged.

That's the notes covered.

Carry on to the next subtopic.

Exam questions on Integration by substitution

Let $I=\int 2x(x^2+3)^4\,dx$ . Use the substitution $u=x^2+3$ .
Hence find the exact value of $\int_0^1 2x(x^2+3)^4\,dx$ .2 marks
Let $I=\int x\sqrt{x-2}\,dx$ . Use the substitution $u=x-2$ .
Hence find the exact value of $\int_2^6x\sqrt{x-2}\,dx$ .2 marks
Let $I=\int_0^4\frac{x}{\sqrt{2x+1}}\,dx$ .
Using the substitution $u=2x+1$ , show that $I=\frac14\int_1^9\left(u^{1/2}-u^{-1/2}\right)du$ .3 marks

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Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).