2.4 - Choosing an integration method | Cambridge A-Level Further Mathematics

Some integrals need a sequence of familiar methods rather than one new formula. You will solve cyclic integrations by parts, turn trigonometric expressions into rational functions using a half-angle substitution, and choose a second method after a first substitution has simplified the problem.

Repeated integration by parts

What should you do if integration by parts brings back the integral you started with? For an exponential multiplied by a sine or cosine, that return is useful: it creates an equation for the unknown integral.

You already know $\int u v'\,\mathrm dx=uv-\int u'v\,\mathrm dx$ . Choosing the exponential as $v'$ keeps it easy to integrate; differentiating sine twice brings it back with a minus sign. That is why two applications of parts can close the calculation.

Problem-solving

Name the integral, apply parts twice and collect the repeated integral on one side. Work with representative antiderivatives during the rearrangement, then include one arbitrary constant in the final answer.

Use different letters for the two trig integrals so that their signs remain visible.

Worked example 1: a cyclic sine integral

Question. Find $\displaystyle\int e^{2x}\sin3x\,\mathrm dx$ .

Plan. Apply parts twice, keeping $e^{2x}/2$ as the integrated exponential, then solve for the original integral.

Let $I=\int e^{2x}\sin3x\,\mathrm dx$ and $J=\int e^{2x}\cos3x\,\mathrm dx$ . Integrating the exponential in each application gives

I=\frac12e^{2x}\sin3x-\frac32J, \qquad J=\frac12e^{2x}\cos3x+\frac32I.

The plus sign in the second relation comes from differentiating cosine to $-3\sin3x$ . Substitute for $J$ :

\begin{aligned} I&=\frac12e^{2x}\sin3x-\frac34e^{2x}\cos3x-\frac94I\\ \frac{13}{4}I&=\frac12e^{2x}\sin3x-\frac34e^{2x}\cos3x \end{aligned}

Answer. $\displaystyle\frac{e^{2x}}{13}(2\sin3x-3\cos3x)+C$ .

Check. Product differentiation gives coefficients $4+9=13$ for sine and $-6+6=0$ for cosine, all divided by $13$ . The derivative is exactly $e^{2x}\sin3x$ .

Change only sine to cosine. The same two relations apply, but solve for $J$ instead. This is a useful way to predict the sign change before calculating.

Worked example 2: the cyclic cosine integral

Question. Find $\displaystyle\int e^{2x}\cos3x\,\mathrm dx$ .

Plan. Apply parts to cosine, substitute the sine relation and collect the repeated cosine integral.

With $J$ now the target,

J=\frac12e^{2x}\cos3x+\frac32I, \qquad I=\frac12e^{2x}\sin3x-\frac32J.

Therefore

\begin{aligned} J&=\frac12e^{2x}\cos3x+\frac34e^{2x}\sin3x-\frac94J\\ \frac{13}{4}J&=\frac12e^{2x}\cos3x+\frac34e^{2x}\sin3x \end{aligned}

Answer. $\displaystyle\frac{e^{2x}}{13}(2\cos3x+3\sin3x)+C$ .

Check. Differentiate the answer. The cosine terms have total coefficient $4+9=13$ , while the sine terms have coefficient $6-6=0$ . This recovers the required integrand. Also, at $x=0$ the derivative is $1$ , matching $e^0\cos0$ .

The denominator $13$ is $2^2+3^2$ , produced by rearranging the cycle. There is no need to memorise an exponential-trigonometric integral formula; neither of these final results is listed in the formula booklet.

Watch out

The return of $I$ does not mean the method has failed. Stop repeating the same calculation and solve the equation. In particular, $I=A-\tfrac94I$ gives $\tfrac{13}{4}I=A$ , not $-\tfrac54I=A$ .

Use the same cycle with new exponential and angle scales.

The tangent half-angle substitution

The substitution $t=\tan(x/2)$ writes sine, cosine and the differential as rational expressions in the same variable. This can turn an integral such as $\int1/(2+\cos x)\,\mathrm dx$ , where no numerator derivative is available, into an ordinary quadratic-denominator integral.

The double-angle identities give

\sin x=\frac{2\tan(x/2)}{1+\tan^2(x/2)}, \qquad \cos x=\frac{1-\tan^2(x/2)}{1+\tan^2(x/2)}.

Also, differentiating $t=\tan(x/2)$ gives $\mathrm dt/\mathrm dx=(1+t^2)/2$ .

Tangent half-angle conversion

For $t=\tan(x/2)$ ,

\sin x=\frac{2t}{1+t^2}, \qquad \cos x=\frac{1-t^2}{1+t^2}, \qquad \mathrm dx=\frac{2\,\mathrm dt}{1+t^2}.

The sine and cosine conversions are in the formula booklet. The differential is not printed with them: derive it by differentiation. The substitution is undefined at odd multiples of $\pi$ ; use it on an interval between such points.

On $-\pi<x<\pi$ , the variable $t$ increases through all real values. The resulting antiderivative below is valid on that interval; a formula involving $\tan(x/2)$ need not itself extend continuously through its endpoints, even if the original integrand does.

Worked example 3: a cosine denominator

Question. Use $t=\tan(x/2)$ to find $\displaystyle\int\frac{\mathrm dx}{2+\cos x}$ , on $-\pi<x<\pi$ .

Plan. Convert both cosine and the differential, cancel $1+t^2$ , then use an inverse tangent integral.

\begin{aligned} \int\frac{\mathrm dx}{2+\cos x} &=\int\frac{2\,\mathrm dt}{2(1+t^2)+1-t^2}\\ &=2\int\frac{\mathrm dt}{t^2+3}\\ &=\frac2{\sqrt3}\tan^{-1}\left(\frac t{\sqrt3}\right)+C \end{aligned}

For the last step, put $v=t/\sqrt3$ and use $\int(1+v^2)^{-1}\,\mathrm dv=\tan^{-1}v+C$ .

Answer. $\displaystyle\frac2{\sqrt3}\tan^{-1}\left(\frac{\tan(x/2)}{\sqrt3}\right)+C$ .

Check. With $t=\tan(x/2)$ , differentiation of the answer gives $(1+t^2)/(3+t^2)$ . Replacing $t^2$ by $(1-\cos x)/(1+\cos x)$ simplifies this to $1/(2+\cos x)$ , verifying the answer directly.

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2.4 - Choosing an integration method

Repeated integration by parts

The tangent half-angle substitution

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