3.5 - Integration by parts | Cambridge A-Level Mathematics

The product of two easy functions may be awkward to integrate. Integration by parts reverses the product rule and replaces one integral with another, chosen to be simpler. In this part of 3.5, you will make that choice, repeat the method when necessary and uncover products hidden inside logarithmic and inverse-tangent integrals.

Reversing the product rule

What differentiates to $xe^{2x}$ ? The tempting answer $xe^{2x}/2$ produces the wanted term, but also an unwanted $e^{2x}/2$ . Integration by parts organises exactly this extra-term correction.

The product rule says

\frac{\mathrm d}{\mathrm dx}(uv) =u\frac{\mathrm dv}{\mathrm dx}+v\frac{\mathrm du}{\mathrm dx}.

Integrating and rearranging gives the new rule. We deliberately choose which factor to differentiate, $u$ , and which factor to integrate, $\mathrm dv/\mathrm dx$ .

Integration by parts

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

Choose $u$ so that differentiating it simplifies the remaining integral, and choose $\mathrm dv/\mathrm dx$ so that you can find $v$ . Include an arbitrary constant in the final indefinite result.

In the formula booklet.

For $xe^{2x}$ , choose $u=x$ because its derivative is $1$ , while integrating the exponential keeps its shape. Taking $u=e^{2x}$ instead would introduce an $x^2$ factor into the remaining integral, making it harder.

Worked example 1: a polynomial and an exponential

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

Plan. Differentiate $x$ , integrate $e^{2x}$ , then subtract the extra integral created by the product rule.

u=x,\quad \frac{\mathrm du}{\mathrm dx}=1, \qquad \frac{\mathrm dv}{\mathrm dx}=e^{2x},\quad v=\frac12e^{2x}.

Thus

\begin{aligned} \int xe^{2x}\,\mathrm dx &=\frac{x}{2}e^{2x}-\frac12\int e^{2x}\,\mathrm dx\\ &=\frac{x}{2}e^{2x}-\frac14e^{2x}+C \end{aligned}

Answer. $e^{2x}(x/2-1/4)+C$ .

Check. Differentiating the expanded answer gives $e^{2x}/2+xe^{2x}-e^{2x}/2=xe^{2x}$ . The unwanted terms cancel.

Keep the factor $x$ and replace the exponential with a sine of the same argument. The choice of $u$ stays the same; the integral of sine introduces a negative sign.

Worked example 2: a polynomial and a sine

Question. Find $\displaystyle\int x\sin(2x)\,\mathrm dx$ .

Plan. Differentiate $x$ , integrate the sine and keep the subtraction in the parts formula visible.

Take $u=x$ , $\mathrm du/\mathrm dx=1$ , $\mathrm dv/\mathrm dx=\sin(2x)$ , $v=-\cos(2x)/2$ . Therefore

\begin{aligned} \int x\sin(2x)\,\mathrm dx &=-\frac{x}{2}\cos(2x)-\int\left(-\frac12\cos(2x)\right)\,\mathrm dx\\ &=-\frac{x}{2}\cos(2x)+\frac14\sin(2x)+C \end{aligned}

Answer. $-x\cos(2x)/2+\sin(2x)/4+C$ , with angles in radians.

Check. Its derivative is $-\cos(2x)/2+x\sin(2x)+\cos(2x)/2=x\sin(2x)$ .

The method does not multiply the separate antiderivatives of the two factors. It replaces the original integral with a product minus another integral, and that second integral is what removes the unwanted derivative term.

Watch out

The formula has a subtraction: $uv-\int vu'\,\mathrm dx$ . If $v$ is already negative, write brackets before simplifying. Dropping either sign spoils the cancellation in the derivative check.

Use the same choice of the factor to differentiate, this time with a cosine.

Repeating the method

A quadratic factor becomes linear after one differentiation, then constant after a second. If one use of integration by parts leaves a polynomial times an exponential or trigonometric function, use it again. The polynomial's falling degree tells you that this process will finish.

For instance, starting with $x^2e^{-x}$ , choosing $u=x^2$ and $v=-e^{-x}$ leaves an integral involving $xe^{-x}$ . That is simpler, even though it still needs integration by parts.

Problem-solving

Give the remaining integral a name if signs become hard to track. Finish it separately, then substitute the whole result back with brackets. A simpler integral need not be immediately integrable in one step.

The first example carries the negative exponential through both applications.

Worked example 3: two applications with a decaying exponential

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

Plan. Differentiate the polynomial each time, reducing its degree from two to one and then zero.

Let $I=\int x^2e^{-x}\,\mathrm dx$ . With $u=x^2$ , $v=-e^{-x}$ ,

I=-x^2e^{-x}+2\int xe^{-x}\,\mathrm dx.

For the remaining integral, use $u=x$ , $v=-e^{-x}$ :

\begin{aligned} \int xe^{-x}\,\mathrm dx &=-xe^{-x}+\int e^{-x}\,\mathrm dx\\ &=-xe^{-x}-e^{-x} \end{aligned}

Using one final arbitrary constant,

I=-x^2e^{-x}-2xe^{-x}-2e^{-x}+C.

Answer. $-e^{-x}(x^2+2x+2)+C$ .

Check. Product differentiation gives $e^{-x}(x^2+2x+2)-e^{-x}(2x+2)=x^2e^{-x}$ .

Keep the quadratic factor and change only the other factor to $\cos x$ . The degree still falls twice; the sine and cosine signs now alternate.

Worked example 4: two applications with cosine

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

Plan. Reduce the polynomial degree twice, keeping the coefficient $-2$ outside the second integral until it is finished.

First choose $u=x^2$ , $v=\sin x$ :

I=x^2\sin x-2\int x\sin x\,\mathrm dx.

In the remaining integral, choose $u=x$ , $v=-\cos x$ :

\int x\sin x\,\mathrm dx=-x\cos x+\sin x.

Therefore

\begin{aligned} I&=x^2\sin x-2(-x\cos x+\sin x)+C\\ &=x^2\sin x+2x\cos x-2\sin x+C \end{aligned}

Answer. $x^2\sin x+2x\cos x-2\sin x+C$ .

Check. Differentiation gives

2x\sin x+x^2\cos x+2\cos x-2x\sin x-2\cos x=x^2\cos x.

Continue this lesson.

You’re previewing NovaLearn without an account. Create a free account to read the rest.

Create free account

Last Accessed

Pure Mathematics 1

Advanced pure mathematics

Pure Mathematics 3 extensions

Pure Mathematics 2 only

Mechanics

Probability and Statistics 1

Probability and Statistics 2

3.5 - Integration by parts

Reversing the product rule

Repeating the method

Continue this lesson.

Learn faster with Nova