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EnglandMathsSyllabus dot point

How do you integrate harder functions using substitution, integration by parts, and partial fractions?

Integration by substitution, integration by parts, integration using partial fractions, and integrating expressions of the form f prime over f and products reducible by a trigonometric identity.

A focused answer to the OCR A-Level Mathematics A advanced integration content, covering integration by substitution, integration by parts, integrating with partial fractions, the f prime over f logarithm pattern, and using trigonometric identities to integrate products such as sine squared.

Generated by Claude Opus 4.812 min answer

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  1. What this dot point is asking
  2. The answer
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What this dot point is asking

OCR wants you to integrate harder expressions using substitution (including changing the limits for a definite integral), integration by parts, and partial fractions, to recognise the f(x)f(x)\dfrac{f'(x)}{f(x)} pattern that integrates to a logarithm, and to use trigonometric identities to integrate squared functions such as sin2x\sin^2 x.

The answer

Integration by substitution

Substitution reverses the chain rule. Pick an inner expression as uu, replace dxdx using du=dudxdxdu = \dfrac{du}{dx}\,dx, rewrite the whole integral in uu, integrate, and (for an indefinite integral) substitute back.

For a definite integral it is cleaner to change the limits to values of uu rather than substituting back.

The logarithm pattern

When the numerator is the derivative of the denominator, the integral is a logarithm. If the numerator is a constant multiple of the derivative, adjust by that constant. This pattern is worth spotting before reaching for a full substitution, because it gives the answer in one line.

Integration by parts

Integration by parts reverses the product rule. Choose uu to be the factor that simplifies when differentiated, and dvdx\dfrac{dv}{dx} to be the factor you can integrate.

A useful guide for choosing uu: logarithms and powers of xx simplify on differentiating, so they make good choices for uu; exponentials and sines stay manageable when integrated, so they make good choices for dvdx\dfrac{dv}{dx}.

Partial fractions

A rational function with a factorised denominator is integrated by splitting it into partial fractions first, after which each piece becomes a logarithm.

Choosing the right technique

With several techniques available, a quick decision tree helps. If you spot f(x)f(x)\dfrac{f'(x)}{f(x)}, write down the logarithm immediately. If the integrand is a product where one factor is the derivative of the inside of the other, use substitution. If it is a product of two unrelated functions (such as xx times an exponential or a trigonometric function), use parts. If it is a rational function, split into partial fractions. If it contains sin2x\sin^2 x or cos2x\cos^2 x, apply a double angle identity first. Recognising the form quickly is worth as much as the technique itself in a timed paper.

Examples in context

A trigonometric identity first

You cannot integrate sin2x\sin^2 x directly, but the double angle identity turns it into something you can.

Integrating with partial fractions

Try this

Q1. Find 6x2x3+2dx\displaystyle\int \dfrac{6x^2}{x^3 + 2}\,dx. [2 marks]

  • Cue. Numerator is twice the derivative of the denominator: 2lnx3+2+c2\ln|x^3 + 2| + c.

Q2. Find xsinxdx\displaystyle\int x\sin x\,dx by parts. [3 marks]

  • Cue. u=xu = x, dvdx=sinx\frac{dv}{dx} = \sin x, so xcosx+cosxdx=xcosx+sinx+c-x\cos x + \int\cos x\,dx = -x\cos x + \sin x + c.

Exam-style practice questions

Practice questions written in the style of OCR exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

OCR 20195 marksUse integration by parts to find xe2xdx\int x e^{2x}\,dx.
Show worked answer →

Choose u=xu = x (simplifies on differentiating) and dvdx=e2x\dfrac{dv}{dx} = e^{2x} (M1), so dudx=1\dfrac{du}{dx} = 1 and v=12e2xv = \tfrac{1}{2}e^{2x} (A1).

Apply udvdxdx=uvvdudxdx\int u\dfrac{dv}{dx}\,dx = uv - \int v\dfrac{du}{dx}\,dx (M1): xe2xdx=12xe2x12e2xdx\int x e^{2x}\,dx = \tfrac{1}{2}x e^{2x} - \int \tfrac{1}{2}e^{2x}\,dx (A1).

The remaining integral gives xe2xdx=12xe2x14e2x+c\int x e^{2x}\,dx = \tfrac{1}{2}x e^{2x} - \tfrac{1}{4}e^{2x} + c (A1).

Markers reward a sensible choice of uu and vv, applying the by-parts formula, integrating the second term, and the constant.

OCR 20216 marksUse the substitution u=x2+1u = x^2 + 1 to find 2xx2+1dx\int 2x\sqrt{x^2 + 1}\,dx, then evaluate 032xx2+1dx\int_0^{\sqrt{3}} 2x\sqrt{x^2 + 1}\,dx.
Show worked answer →

With u=x2+1u = x^2 + 1, dudx=2x\dfrac{du}{dx} = 2x, so du=2xdxdu = 2x\,dx (M1). The integral becomes udu\int \sqrt{u}\,du (A1).

Integrate: u1/2du=23u3/2+c=23(x2+1)3/2+c\int u^{1/2}\,du = \tfrac{2}{3}u^{3/2} + c = \tfrac{2}{3}(x^2 + 1)^{3/2} + c (A1).

For the definite integral, change the limits: x=0u=1x = 0 \Rightarrow u = 1 and x=3u=4x = \sqrt{3} \Rightarrow u = 4 (M1).

[23u3/2]14=23(8)23(1)=16323=143\big[\tfrac{2}{3}u^{3/2}\big]_1^4 = \tfrac{2}{3}(8) - \tfrac{2}{3}(1) = \tfrac{16}{3} - \tfrac{2}{3} = \tfrac{14}{3} (M1, A1).

Markers reward the substitution and dudu, the integral in uu, changing the limits, and the final value.

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