Year 2 Pure · Chapter 15

Integration Techniques

Specification: Edexcel Pure 8.2, 8.5, 8.6 — Integration · AQA section H2, H5, H6

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In this chapter

1. Standard integrals
2. Recognising a reversed chain rule
3. Integration by substitution
4. Integration by parts
5. Using partial fractions
Practice

Key facts — this chapter

$\displaystyle\int e^{kx}\,dx = \tfrac{1}{k}e^{kx} + c$ , $\displaystyle\int \frac{1}{x}\,dx = \ln|x| + c$ , $\displaystyle\int \cos kx\,dx = \tfrac{1}{k}\sin kx + c$ and $\displaystyle\int \sin kx\,dx = -\tfrac{1}{k}\cos kx + c$ .
Reverse chain rule: $\displaystyle\int \frac{f'(x)}{f(x)}\,dx = \ln|f(x)| + c$ , and $\displaystyle\int f'(x)\big(f(x)\big)^n\,dx = \frac{\big(f(x)\big)^{n+1}}{n + 1} + c$ .
Use identities to turn $\sin^2 x$ , $\cos^2 x$ and $\tan^2 x$ into things you can integrate.
Substitution: replace $x$ by $u$ , $dx$ by $\frac{dx}{du}\,du$ , and change the limits.
By parts: $\displaystyle\int u\frac{dv}{dx}\,dx = uv - \int v\frac{du}{dx}\,dx$ . Choose $u$ to be the factor that gets simpler when differentiated, except that $\ln x$ is always $u$ .
Partial fractions split a fraction into pieces that each integrate to a logarithm.

Year 1 integrated powers of $x$ . This chapter reverses every derivative from Chapter 12, then adds methods for the integrals that don’t match anything directly: spotting a reversed chain rule, substituting a new variable, integrating a product by parts, and splitting a fraction.

1. Standard integrals

Each derivative from Chapter 12, read backwards, gives an integral.

Key result

\begin{aligned} \int e^{kx}\,dx &= \tfrac{1}{k}e^{kx} + c \\ \int \frac{1}{x}\,dx &= \ln|x| + c \\ \int \cos kx\,dx &= \tfrac{1}{k}\sin kx + c \\ \int \sin kx\,dx &= -\tfrac{1}{k}\cos kx + c \\ \int \sec^2 kx\,dx &= \tfrac{1}{k}\tan kx + c \end{aligned}

More generally, if $\int f(x)\,dx = F(x) + c$ , then $\int f(ax + b)\,dx = \frac{1}{a}F(ax + b) + c$ .

Worked example 15.1 · A sum of standard integrals

Find $\displaystyle\int \left(e^{3x} + \frac{4}{x} - 2\sin 5x\right)dx$ .

Integrate term by term:

\tfrac{1}{3}e^{3x} + 4\ln|x| + \tfrac{2}{5}\cos 5x + c

Worked example 15.2 · A definite integral

Find the exact value of $\displaystyle\int_0^{\pi/6} \sec^2 2x\,dx$ .

\begin{aligned} \int_0^{\pi/6} \sec^2 2x\,dx &= \Big[\tfrac{1}{2}\tan 2x\Big]_0^{\pi/6} \\ &= \tfrac{1}{2}\tan\tfrac{\pi}{3} - 0 = \tfrac{\sqrt3}{2} \end{aligned}

Common mistake

The integral of $\frac{1}{x}$ is $\ln|x|$ , with the modulus, because $\ln x$ is only defined for $x > 0$ . So $\int_{-3}^{-1} \frac{1}{x}\,dx = \ln 1 - \ln 3 = -\ln 3$ , which is negative, as it should be for an area below the axis.

2. Recognising a reversed chain rule

Differentiating $\ln|f(x)|$ or $\big(f(x)\big)^{n+1}$ by the chain rule gives a recognisable pattern: something multiplied by the derivative of its inside. When an integral has that pattern, you can reverse the chain rule directly.

Key result

\begin{aligned} \int \frac{f'(x)}{f(x)}\,dx &= \ln|f(x)| + c \\ \int f'(x)\big(f(x)\big)^n\,dx &= \frac{\big(f(x)\big)^{n+1}}{n + 1} + c \end{aligned}

The second works for any $n \ne -1$ . If the top is only a multiple of $f'(x)$ , adjust by that constant.

Worked example 15.3 · The top is the derivative of the bottom

Find (a) $\displaystyle\int \frac{2x}{x^2 + 1}\,dx$ and (b) $\displaystyle\int \tan x\,dx$ .

(a) The top is the derivative of $x^2 + 1$ , so the integral is $\ln(x^2 + 1) + c$ . No modulus is needed, because $x^2 + 1 > 0$ .

(b) $\tan x = \dfrac{\sin x}{\cos x}$ , and $\sin x$ is minus the derivative of $\cos x$ . So

\begin{aligned} \int \tan x\,dx &= -\ln|\cos x| + c \\ &= \ln|\sec x| + c \end{aligned}

Worked example 15.4 · A power of a bracket

Find $\displaystyle\int x(x^2 + 3)^4\,dx$ .

The derivative of $x^2 + 3$ is $2x$ , and the integral has $x$ , which is half of it. So try $(x^2 + 3)^5$ : its derivative is $10x(x^2 + 3)^4$ , which is ten times too big. Adjust:

\int x(x^2 + 3)^4\,dx = \tfrac{1}{10}(x^2 + 3)^5 + c

Using trigonometric identities

$\sin^2 x$ and $\cos^2 x$ have no standard integral, but the double-angle formulae of Chapter 10 turn them into ones that do. $\tan^2 x$ becomes $\sec^2 x - 1$ .

Key result

\begin{aligned} \sin^2 x &= \tfrac{1}{2}(1 - \cos 2x) \\ \cos^2 x &= \tfrac{1}{2}(1 + \cos 2x) \\ \tan^2 x &= \sec^2 x - 1 \end{aligned}

Worked example 15.5 · Squares of trig functions

Find $\displaystyle\int \sin^2 x\,dx$ , and the exact value of $\displaystyle\int_0^{\pi/4} \tan^2 x\,dx$ .

$\displaystyle\int \tfrac{1}{2}(1 - \cos 2x)\,dx = \tfrac{1}{2}x - \tfrac{1}{4}\sin 2x + c$ .
For the second:
$\begin{aligned} &\int_0^{\pi/4} (\sec^2 x - 1)\,dx \\ &= \Big[\tan x - x\Big]_0^{\pi/4} = 1 - \tfrac{\pi}{4} \end{aligned}$

3. Integration by substitution

When the pattern isn’t obvious, change the variable. Choose $u$ as some function of $x$ , then rewrite everything in terms of $u$ , including $dx$ .

Key result

To substitute $u = g(x)$ :

Find $\frac{du}{dx}$ and use it to replace $dx$ with $\frac{dx}{du}\,du$ .
Write the rest of the integrand in terms of $u$ .
For a definite integral, change the limits to values of $u$ .
Integrate, and for an indefinite integral, put $x$ back at the end.

Worked example 15.6 · An indefinite substitution

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

$\frac{du}{dx} = 2$ , so $dx = \tfrac{1}{2}\,du$ , and $x = \tfrac{u - 1}{2}$ .
Substitute and integrate:
$\begin{aligned} &\int \frac{u - 1}{2}\sqrt{u} \times \frac{1}{2}\,du \\ &= \tfrac{1}{4}\int \left(u^{\frac{3}{2}} - u^{\frac{1}{2}}\right)du \\ &= \tfrac{1}{10}u^{\frac{5}{2}} - \tfrac{1}{6}u^{\frac{3}{2}} + c \end{aligned}$
Put $x$ back: $\tfrac{1}{10}(2x + 1)^{\frac{5}{2}} - \tfrac{1}{6}(2x + 1)^{\frac{3}{2}} + c$ .

Worked example 15.7 · Changing the limits

Use the substitution $u = x + 1$ to find $\displaystyle\int_0^3 \frac{x}{\sqrt{x + 1}}\,dx$ .

$dx = du$ and $x = u - 1$ . When $x = 0$ , $u = 1$ , and when $x = 3$ , $u = 4$ .
So the integral becomes
$\begin{aligned} &\int_1^4 \frac{u - 1}{\sqrt u}\,du \\ &= \int_1^4 \left(u^{\frac{1}{2}} - u^{-\frac{1}{2}}\right)du \\ &= \Big[\tfrac{2}{3}u^{\frac{3}{2}} - 2u^{\frac{1}{2}}\Big]_1^4 \\ &= \left(\tfrac{16}{3} - 4\right) - \left(\tfrac{2}{3} - 2\right) = \tfrac{8}{3} \end{aligned}$

With the limits changed, there’s no need to go back to $x$ .

Left: the area under y = x over the square root of (x + 1) from x = 0 to 3, shaded. Right: the area under y = (u minus 1) over the square root of u from u = 1 to 4, shaded. Both areas are 8 over 3 — The substitution u = x + 1 turns the area on the left into the one on the right. They are equal: both are 8/3.

The left panel is an integral in x; the right is the same integral after the substitution. Move the upper limit and compare the shaded areas: as x runs over its interval, u runs over the new one, and the two areas stay equal.

4. Integration by parts

Integrating the product rule gives a way to integrate some products.

Key result

\int u\frac{dv}{dx}\,dx = uv - \int v\frac{du}{dx}\,dx

Choose $u$ to be the factor that gets simpler when you differentiate it, such as a power of $x$ , so the new integral is easier. The exception is $\ln x$ : it has no standard integral, so always make it $u$ .

Where it comes from

The product rule says $\frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx}$ . Integrate both sides with respect to $x$ : $uv = \int u\frac{dv}{dx}\,dx + \int v\frac{du}{dx}\,dx$ . Rearranging gives the formula. The figure shows the same thing with areas.

A curve in the (u, v) plane from the point (u1, v1) to (u2, v2). The area between the curve and the u axis is the integral of v du, and the area between the curve and the v axis is the integral of u dv. Together with the small rectangle u1 v1 they make the big rectangle u2 v2 — The two integrals are the two areas either side of the curve. Together they fill the difference between the rectangles, u₂v₂ − u₁v₁.

Worked example 15.8 · A power times an exponential

Find $\displaystyle\int xe^{2x}\,dx$ .

Let $u = x$ , which gets simpler, and $\frac{dv}{dx} = e^{2x}$ . Then $\frac{du}{dx} = 1$ and $v = \tfrac{1}{2}e^{2x}$ .
By parts:
$\begin{aligned} &\int xe^{2x}\,dx \\ &= \tfrac{1}{2}xe^{2x} - \int \tfrac{1}{2}e^{2x}\,dx \\ &= \tfrac{1}{2}xe^{2x} - \tfrac{1}{4}e^{2x} + c \end{aligned}$

Worked example 15.9 · Integrating ln x

Find $\displaystyle\int \ln x\,dx$ , and hence the exact value of $\displaystyle\int_1^3 \ln x\,dx$ .

Write $\ln x$ as $\ln x \times 1$ . Let $u = \ln x$ and $\frac{dv}{dx} = 1$ , so $\frac{du}{dx} = \frac{1}{x}$ and $v = x$ :
$\begin{aligned} &\int \ln x\,dx \\ &= x\ln x - \int x \times \frac{1}{x}\,dx \\ &= x\ln x - x + c \end{aligned}$
So $\displaystyle\int_1^3 \ln x\,dx = (3\ln 3 - 3) - (0 - 1) = 3\ln 3 - 2$ . This is the exact value the trapezium rule estimated in Chapter 14.

Common mistake

Some integrals need parts twice. For $\int x^2e^x\,dx$ , the first use leaves $\int 2xe^x\,dx$ , which needs parts again with $u = 2x$ . Keep choosing the power of $x$ as $u$ each time, or the integral gets worse instead of better.

5. Using partial fractions

A fraction whose bottom factorises can be split into partial fractions (Chapter 2). Each piece with a linear bottom integrates to a logarithm.

Key result

\int \frac{A}{ax + b}\,dx = \frac{A}{a}\ln|ax + b| + c

Worked example 15.10 · A definite integral with partial fractions

Find the exact value of $\displaystyle\int_2^3 \frac{1}{x^2 - 1}\,dx$ .

$x^2 - 1 = (x - 1)(x + 1)$ , and the partial fractions are
$\frac{1}{x^2 - 1} = \frac{1}{2}\left(\frac{1}{x - 1} - \frac{1}{x + 1}\right)$
Integrate and use the laws of logarithms:
$\begin{aligned} &\tfrac{1}{2}\Big[\ln|x - 1| - \ln|x + 1|\Big]_2^3 \\ &= \tfrac{1}{2}\big[(\ln 2 - \ln 4) - (\ln 1 - \ln 3)\big] \\ &= \tfrac{1}{2}\ln\tfrac{3}{2} \end{aligned}$

Practice

Twelve questions in three tiers. Try each one properly before opening the solution — the solutions show full method, not just answers.

Warm-up

1 [4 marks]

Find each of these.

$\displaystyle\int e^{4x}\,dx$ [1]
$\displaystyle\int \frac{3}{x}\,dx$ [1]
$\displaystyle\int \cos 2x\,dx$ [1]
$\displaystyle\int \sec^2 3x\,dx$ [1]

Show solution

(a) $\tfrac{1}{4}e^{4x} + c$ (b) $3\ln|x| + c$ (c) $\tfrac{1}{2}\sin 2x + c$ (d) $\tfrac{1}{3}\tan 3x + c$

2 [2 marks]

Find the exact value of $\displaystyle\int_0^{\pi/2} \sin 2x\,dx$ .

Show solution

\Big[-\tfrac{1}{2}\cos 2x\Big]_0^{\pi/2} = \tfrac{1}{2} - \left(-\tfrac{1}{2}\right) = 1

3 [2 marks]

Find $\displaystyle\int \frac{6x^2}{x^3 + 1}\,dx$ .

Show solution

The top is twice the derivative of $x^3 + 1$ , so the integral is $2\ln|x^3 + 1| + c$ .

4 [2 marks]

Find $\displaystyle\int \cos^2 x\,dx$ .

Show solution

\begin{aligned} &\int \tfrac{1}{2}(1 + \cos 2x)\,dx \\ &= \tfrac{1}{2}x + \tfrac{1}{4}\sin 2x + c \end{aligned}

Standard

5 [3 marks]

Find $\displaystyle\int x\cos x\,dx$ .

Show solution

By parts with $u = x$ and $\frac{dv}{dx} = \cos x$ , so $v = \sin x$ :

\begin{aligned} \int x\cos x\,dx &= x\sin x - \int \sin x\,dx \\ &= x\sin x + \cos x + c \end{aligned}

6 [4 marks]

Use the substitution $u = x - 2$ to find $\displaystyle\int x(x - 2)^6\,dx$ .

Show solution

$dx = du$ and $x = u + 2$ , so

\begin{aligned} &\int (u + 2)u^6\,du = \int (u^7 + 2u^6)\,du \\ &= \tfrac{1}{8}u^8 + \tfrac{2}{7}u^7 + c \\ &= \tfrac{1}{8}(x - 2)^8 + \tfrac{2}{7}(x - 2)^7 + c \end{aligned}

7 [5 marks]

Find the exact value of $\displaystyle\int_1^e x\ln x\,dx$ .

Show solution

$\ln x$ is $u$ , and $\frac{dv}{dx} = x$ , so $v = \tfrac{1}{2}x^2$ :

\begin{aligned} \int x\ln x\,dx &= \tfrac{1}{2}x^2\ln x - \int \tfrac{1}{2}x\,dx \\ &= \tfrac{1}{2}x^2\ln x - \tfrac{1}{4}x^2 + c \end{aligned}

So the definite integral is

\left(\tfrac{1}{2}e^2 - \tfrac{1}{4}e^2\right) - \left(0 - \tfrac{1}{4}\right) = \tfrac{1}{4}(e^2 + 1)

8 [5 marks]

Express $\dfrac{x + 5}{(x + 1)(x - 3)}$ in partial fractions. [3]
Hence find $\displaystyle\int \frac{x + 5}{(x + 1)(x - 3)}\,dx$ . [2]

Show solution

(a) $x + 5 \equiv A(x - 3) + B(x + 1)$ . Put $x = 3$ : $8 = 4B$ , so $B = 2$ . Put $x = -1$ : $4 = -4A$ , so $A = -1$ . So the fraction is $\dfrac{2}{x - 3} - \dfrac{1}{x + 1}$ .

(b) $2\ln|x - 3| - \ln|x + 1| + c$ .

9 [3 marks]

Find the exact value of $\displaystyle\int_0^{\pi/4} \sin x\cos x\,dx$ .

Show solution

$\sin x\cos x = \tfrac{1}{2}\sin 2x$ , so

\begin{aligned} &\int_0^{\pi/4} \tfrac{1}{2}\sin 2x\,dx \\ &= \Big[-\tfrac{1}{4}\cos 2x\Big]_0^{\pi/4} = 0 + \tfrac{1}{4} = \tfrac{1}{4} \end{aligned}

Exam style

10 [6 marks]

Use the substitution $u = 2x - 1$ to find the exact value of $\displaystyle\int_1^5 \frac{x}{\sqrt{2x - 1}}\,dx$ .

Show solution

$\frac{du}{dx} = 2$ , so $dx = \tfrac{1}{2}\,du$ , and $x = \tfrac{u + 1}{2}$ . When $x = 1$ , $u = 1$ , and when $x = 5$ , $u = 9$ . So the integral is

\begin{aligned} &\int_1^9 \frac{u + 1}{2\sqrt u} \times \frac{1}{2}\,du \\ &= \tfrac{1}{4}\int_1^9 \left(u^{\frac{1}{2}} + u^{-\frac{1}{2}}\right)du \\ &= \tfrac{1}{4}\Big[\tfrac{2}{3}u^{\frac{3}{2}} + 2u^{\frac{1}{2}}\Big]_1^9 \\ &= \tfrac{1}{4}\left[(18 + 6) - \left(\tfrac{2}{3} + 2\right)\right] = \tfrac{16}{3} \end{aligned}

11 [6 marks]

Find $\displaystyle\int x^2e^x\,dx$ . [4]
Hence find the exact value of $\displaystyle\int_0^1 x^2e^x\,dx$ . [2]

Show solution

(a) By parts with $u = x^2$ and $v = e^x$ , then by parts again on the new integral, with $u = 2x$ :

\begin{aligned} &\int x^2e^x\,dx \\ &= x^2e^x - \int 2xe^x\,dx \\ &= x^2e^x - \left(2xe^x - \int 2e^x\,dx\right) \\ &= e^x(x^2 - 2x + 2) + c \end{aligned}

(b) $e(1 - 2 + 2) - e^0(0 - 0 + 2) = e - 2$ .

12 [7 marks]

Express $\dfrac{3}{(x + 1)(x - 2)}$ in partial fractions. [3]
Hence show that $\displaystyle\int_3^5 \frac{3}{(x + 1)(x - 2)}\,dx = \ln 2$ . [4]

Show solution

(a) $3 \equiv A(x - 2) + B(x + 1)$ . Put $x = 2$ : $3 = 3B$ , so $B = 1$ . Put $x = -1$ : $3 = -3A$ , so $A = -1$ . The fraction is $\dfrac{1}{x - 2} - \dfrac{1}{x + 1}$ .

(b)

\begin{aligned} &\Big[\ln|x - 2| - \ln|x + 1|\Big]_3^5 \\ &= (\ln 3 - \ln 6) - (\ln 1 - \ln 4) \\ &= \ln\frac{3 \times 4}{6} = \ln 2 \end{aligned}