Integration by substitution, by parts and with partial fractions – A-level Maths exam questions with mark scheme

1CoreCalculator4 marks

Find

(a) $\displaystyle\int(2x+3)^{5}\,\mathrm{d}x$ [2]

(b) $\displaystyle\int\left(4\mathrm{e}^{2x}+\frac{3}{x}\right)\mathrm{d}x,\quad x>0$ [2]

2StandardCalculator5 marks

Use integration by parts to find the exact value of $$\int_{0}^{\frac{\pi}{4}}x\cos2x\,\mathrm{d}x$$ [5]

Parts the right way round: $Ax\sin2x-\displaystyle\int B\sin2x\,\mathrm{d}x$M1
$\tfrac12x\sin2x-\displaystyle\int\tfrac12\sin2x\,\mathrm{d}x$A1
$\tfrac12x\sin2x+\tfrac14\cos2x$A1
Substitutes limits $\tfrac{\pi}{4}$ and $0$ into an expression of the correct form and subtractsM1
$\dfrac{\pi}{8}-\dfrac14$A1
Answer: $\dfrac{\pi}{8}-\dfrac14$
Examiner tip: Don't assume the bottom limit gives zero: at $x=0$, $\tfrac14\cos0=\tfrac14$, which must be subtracted.

3StandardCalculator6 marks

Use the substitution $u=2x-1$ to find the exact value of $$\int_{1}^{5}\frac{x}{\sqrt{2x-1}}\,\mathrm{d}x$$ [6]

$\mathrm{d}x=\tfrac12\,\mathrm{d}u$ and $x=\dfrac{u+1}{2}$B1
Fully substitutes to obtain an integral in $u$ only: $\displaystyle\int\frac{u+1}{2\sqrt u}\cdot\frac12\,\mathrm{d}u$M1
$\displaystyle\tfrac14\int\left(u^{\frac12}+u^{-\frac12}\right)\mathrm{d}u$A1
Integrates: $u^{n}\to u^{n+1}$ for at least one term, giving $\tfrac14\left(\tfrac23u^{\frac32}+2u^{\frac12}\right)$M1
Uses limits $u=9$ and $u=1$ (or reverts to $x$ and uses 5 and 1)M1
$\dfrac{16}{3}$A1
Answer: $\dfrac{16}{3}$
Examiner tip: Change the limits to $u$-values (1 and 9) – using the $x$-limits with a $u$-integral is a very common lost mark.

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4StandardCalculator6 marks

(a) Express $\dfrac{x+5}{(x-1)(x+2)}$ in partial fractions. [2]

(b) Hence show that $$\int_{2}^{5}\frac{x+5}{(x-1)(x+2)}\,\mathrm{d}x=\ln\frac{64}{7}$$ [4]

(a) $x+5\equiv A(x+2)+B(x-1)$ and substitutes a value or compares coefficientsM1
$\dfrac{2}{x-1}-\dfrac{1}{x+2}$A1
(b) Integrates to $p\ln(x-1)+q\ln(x+2)$M1
$2\ln(x-1)-\ln(x+2)$A1
Substitutes limits 5 and 2 and subtracts: $(2\ln4-\ln7)-(2\ln1-\ln4)$M1
Uses log laws correctly: $3\ln4-\ln7=\ln64-\ln7=\ln\dfrac{64}{7}$ csoA1*
Answer: (a) $\dfrac{2}{x-1}-\dfrac{1}{x+2}$
Examiner tip: In a 'show that' the log-law steps must be seen: write $3\ln4=\ln64$ before combining.

5StretchCalculator6 marks

$$I=\int_{0}^{\pi}\mathrm{e}^{x}\sin x\,\mathrm{d}x$$

By using integration by parts twice, show that $$I=\frac{1}{2}\left(\mathrm{e}^{\pi}+1\right)$$ [6]

First application of parts: $\pm\mathrm{e}^{x}\cos x\pm\displaystyle\int\mathrm{e}^{x}\cos x\,\mathrm{d}x$ (or, taking $u=\sin x$, $\pm\mathrm{e}^{x}\sin x\pm\displaystyle\int\mathrm{e}^{x}\cos x\,\mathrm{d}x$)M1
$-\mathrm{e}^{x}\cos x+\displaystyle\int\mathrm{e}^{x}\cos x\,\mathrm{d}x$ (or $\mathrm{e}^{x}\sin x-\displaystyle\int\mathrm{e}^{x}\cos x\,\mathrm{d}x$ if $u=\sin x$)A1
Second application of parts with the same choice of $u$: $\displaystyle\int\mathrm{e}^{x}\cos x\,\mathrm{d}x=\mathrm{e}^{x}\sin x-\int\mathrm{e}^{x}\sin x\,\mathrm{d}x$ (or, with $u$ trig, $\mathrm{e}^{x}\cos x+\displaystyle\int\mathrm{e}^{x}\sin x\,\mathrm{d}x$)M1
Recognises the original integral and rearranges: $2I=\left[\mathrm{e}^{x}\sin x-\mathrm{e}^{x}\cos x\right]_{0}^{\pi}$M1
Correct limits: $(0+\mathrm{e}^{\pi})-(0-1)$A1
$I=\tfrac12(\mathrm{e}^{\pi}+1)$ csoA1*
Answer: Proof: $2I=\left[\mathrm{e}^{x}\sin x-\mathrm{e}^{x}\cos x\right]_{0}^{\pi}=\mathrm{e}^{\pi}+1$, so $I=\tfrac12(\mathrm{e}^{\pi}+1)$
Examiner tip: Use the same choice of $u$ (or $\mathrm{d}v/\mathrm{d}x$) both times – swapping just undoes the first step.

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