1.08i - Integration by parts | OCR A-Level Mathematics A

Reverse the product rule to integrate products that substitution does not simplify. Choose factors purposefully, repeat the method when needed, handle logarithms and keep both endpoints in definite integrals.

Reversing the product rule

What correction turns $x\mathrm e^{2x}/2$ into an antiderivative of $x\mathrm e^{2x}$ ? Differentiate it: the product rule produces the wanted term and an unwanted $\mathrm e^{2x}/2$ . Integration by parts removes that extra piece.

You already know

The product rule gives $(uv)'=u'v+uv'$ . Integrating a product does not mean multiplying the separate integrals.

Integrate the product-rule identity and rearrange; its minus sign comes from moving one integral to the other side.

Integration by parts

For suitable differentiable functions on the interval,

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

In the formula sheet.

Choose $u$ to simplify when differentiated, and $v'$ to be easy to integrate. Do not add a constant to intermediate $v$ : it cancels between the two terms. Put one $C$ in the final indefinite answer.

Worked example 1: lowering a polynomial factor

Question. Find $\int x\mathrm e^x\,\mathrm dx$ .

Plan. Take $u=x$ , $u'=1$ , $v'=\mathrm e^x$ , $v=\mathrm e^x$ .

\int x\mathrm e^x\,\mathrm dx =x\mathrm e^x-\int\mathrm e^x\,\mathrm dx =\mathrm e^x(x-1)+C.

Why differentiate $x$ ? It disappears from the remaining integral. The reverse choice would introduce $x^2$ , making the problem harder.

Answer. $\mathrm e^x(x-1)+C$ .

Check. The derivative is $\mathrm e^x(x-1)+\mathrm e^x=x\mathrm e^x$ .

Change only the exponent's coefficient from $1$ to $2$ . Both integrations of the exponential need the new scale factor.

Worked example 2: scaling the exponential

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

Plan. Keep $u=x$ , now with $v=\mathrm e^{2x}/2$ .

\begin{aligned} I&=\frac{x\mathrm e^{2x}}2-\frac12\int\mathrm e^{2x}\,\mathrm dx &&\text{apply parts}\\ &=\mathrm e^{2x}\left(\frac x2-\frac14\right)+C &&\text{integrate again} \end{aligned}

Why another half? The remaining exponential still has inner derivative $2$ .

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

Check. Product-rule terms not containing $x$ cancel, leaving $x\mathrm e^{2x}$ .

For comparison, the trigonometric product $\int x\cos x\,\mathrm dx=x\sin x+\cos x+C$ uses the same choice of polynomial factor, with angles in radians.

Watch out

Wrong: $\int uv'\,\mathrm dx=uv+\int vu'\,\mathrm dx$ . The correct formula subtracts the remaining integral; differentiate to check the cancellation.

Change the exponential's coefficient once more.

Reducing a polynomial factor more than once

A quadratic factor takes two differentiations to become constant. Apply parts twice, keeping the coefficient and minus sign outside the remaining integral visible.

Worked example 3: two applications

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

Plan. Differentiate the polynomial factor at each application.

\begin{aligned} I&=x^2\mathrm e^x-2\int x\mathrm e^x\,\mathrm dx &&\text{first application}\\ \int x\mathrm e^x\,\mathrm dx&=x\mathrm e^x-\mathrm e^x &&\text{second application}\\ I&=\mathrm e^x(x^2-2x+2)+C &&\text{substitute and collect} \end{aligned}

Why retain the outer $-2$ ? It multiplies both terms of the second result.

Answer. $\mathrm e^x(x^2-2x+2)+C$ .

Check. Differentiate: $\mathrm e^x[(x^2-2x+2)+(2x-2)]=x^2\mathrm e^x$ .

Change only the exponential's coefficient to $2$ . The polynomial still reduces in two steps.

Worked example 4: repeated scaling

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

Plan. Use $v=\mathrm e^{2x}/2$ at each parts step.

\begin{aligned} I&=\frac{x^2\mathrm e^{2x}}2-\int x\mathrm e^{2x}\,\mathrm dx\\ &=\mathrm e^{2x}\left(\frac{x^2}{2}-\frac x2+\frac14\right)+C. \end{aligned}

Why is the remaining coefficient now $-1$ ? The derivative $2x$ cancels the half in $v$ .

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

Check. Differentiating the bracket and exponential leaves precisely $x^2\mathrm e^{2x}$ .

Trigonometric factors work too. Two parts steps give

\begin{aligned} \int x^2\sin x\,\mathrm dx &=-x^2\cos x+2\int x\cos x\,\mathrm dx\\ &=-x^2\cos x+2x\sin x+2\cos x+C. \end{aligned}

These are finite calculations for specified integrands. Reduction formulae for indexed families are excluded; the displayed results are derived from the supplied parts formula, not separately printed.

Watch out

Wrong: substitute only the first term of the second antiderivative. Right: bracket its whole result before distributing the outer coefficient.

Use two parts steps with the cosine instead.

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1.08i - Integration by parts

Reversing the product rule

Reducing a polynomial factor more than once

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