Integration by Substitution

A2 · P3 · 2 min

Substitution changes the variable of integration to make the integrand simpler. In P3 the substitution is always given in the question; your job is to carry it out correctly, including changing $dx$ and, for a definite integral, the limits.

The method

Method

Write down $u$ in terms of $x$ (given), and find $\dfrac{du}{dx}$ .
Rearrange to express $dx$ in terms of $du$ : $dx = \dfrac{du}{du/dx}$ .
Replace every $x$ in the integrand, including any left over, using the substitution.
For a definite integral, convert the limits to $u$ -values. For an indefinite integral, integrate and then substitute back.

A linear substitution with leftover x

Use $u = 2x + 1$ to find $\displaystyle\int x\sqrt{2x + 1} \, dx$ .

Solution

$\dfrac{du}{dx} = 2$ , so $dx = \tfrac{1}{2}du$ , and $x = \dfrac{u - 1}{2}$ .

\int \frac{u - 1}{2}\sqrt{u} \cdot \frac{1}{2} \, du = \frac{1}{4}\int \left(u^{3/2} - u^{1/2}\right) du = \frac{1}{4}\left(\frac{2}{5}u^{5/2} - \frac{2}{3}u^{3/2}\right) + c = \frac{1}{10}(2x + 1)^{5/2} - \frac{1}{6}(2x + 1)^{3/2} + c.

A trigonometric substitution, definite

Use $u = \sin x$ to evaluate $\displaystyle\int_0^{\pi/2} \sin^2 2x\cos x \, dx$ .

Solution

$\sin 2x = 2\sin x\cos x$ , so $\sin^2 2x\cos x = 4\sin^2 x\cos^3 x = 4\sin^2 x(1 - \sin^2 x)\cos x$ .

With $u = \sin x$ , $du = \cos x \, dx$ ; limits $x = 0 \to u = 0$ , $x = \tfrac{\pi}{2} \to u = 1$ .

\int_0^1 4u^2(1 - u^2) \, du = 4\Big[\tfrac{1}{3}u^3 - \tfrac{1}{5}u^5\Big]_0^1 = 4\left(\tfrac{1}{3} - \tfrac{1}{5}\right) = \tfrac{8}{15}.

Substitution producing a log

Use $u = 1 + e^x$ to find $\displaystyle\int \frac{e^{2x}}{1 + e^x} \, dx$ .

Solution

$du = e^x \, dx$ and $e^x = u - 1$ . Then $e^{2x} \, dx = e^x \cdot e^x \, dx = (u - 1) \, du$ .

\int \frac{u - 1}{u} \, du = \int \left(1 - \frac{1}{u}\right) du = u - \ln u + c = e^x - \ln(1 + e^x) + c'.

(The constant $1$ from $u = 1 + e^x$ is absorbed into $c'$ .)

Substitution with x = ... form

Use $x = 2\sin\theta$ to evaluate $\displaystyle\int_0^1 \sqrt{4 - x^2} \, dx$ .

Solution

$dx = 2\cos\theta \, d\theta$ and $\sqrt{4 - x^2} = \sqrt{4 - 4\sin^2\theta} = 2\cos\theta$ . Limits: $x = 0 \to \theta = 0$ , $x = 1 \to \theta = \tfrac{\pi}{6}$ .

\int_0^{\pi/6} 2\cos\theta \cdot 2\cos\theta \, d\theta = 4\int_0^{\pi/6} \cos^2\theta \, d\theta = 2\int_0^{\pi/6} (1 + \cos 2\theta) \, d\theta = 2\Big[\theta + \tfrac{1}{2}\sin 2\theta\Big]_0^{\pi/6} = \tfrac{\pi}{3} + \tfrac{\sqrt{3}}{2}.

Substitution into a log integrand

Use $u = \ln x$ to find $\displaystyle\int \frac{(\ln x)^2}{x} \, dx$ .

Solution

$du = \dfrac{1}{x} \, dx$ , so the integral is $\displaystyle\int u^2 \, du = \tfrac{1}{3}u^3 + c = \tfrac{1}{3}(\ln x)^3 + c$ .

Watch out

Every $x$ must go, including the $dx$ . An integral with a mixture of $x$ and $u$ cannot be evaluated. And when the limits are changed, do not substitute back into $x$ at the end.

Exam tip

Show the three ingredients explicitly: the expression for $dx$ , the transformed integrand, and the new limits. Each usually carries a method mark, and they are the marks you keep even if the final integration slips.

Practice

Question

Use $u = x^2 + 1$ to find $\displaystyle\int \frac{x}{(x^2 + 1)^3} \, dx$ .
Use $u = 3x - 2$ to find $\displaystyle\int \frac{x}{\sqrt{3x - 2}} \, dx$ .
Use $u = \cos x$ to evaluate $\displaystyle\int_0^{\pi/3} \sin x\cos^2 x \, dx$ .
Use $u = \sqrt{x}$ to evaluate $\displaystyle\int_1^4 \frac{1}{\sqrt{x}(1 + \sqrt{x})} \, dx$ .

Answers

$-\dfrac{1}{4(x^2 + 1)^2} + c$ .
$\tfrac{2}{27}(3x - 2)^{3/2} + \tfrac{4}{9}(3x - 2)^{1/2} + c$ .
$\Big[-\tfrac{1}{3}u^3\Big]_1^{1/2} = \tfrac{1}{3}\left(1 - \tfrac{1}{8}\right) = \tfrac{7}{24}$ .
$dx = 2u \, du$ : $\displaystyle\int_1^2 \frac{2}{1 + u} \, du = 2\ln\tfrac{3}{2}$ .

The method

Practice

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