| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2016 |
| Session | June |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration by Substitution |
| Type | Show integral transforms via substitution then evaluate (trigonometric/Weierstrass) |
| Difficulty | Standard +0.8 This is a multi-part question requiring partial fractions (routine C4), then a non-trivial trigonometric substitution with careful algebraic manipulation to reach the target form, followed by integration of sin²θ using double angle formulas. The substitution step requires insight beyond standard textbook exercises, and the multi-stage nature with exact answer requirements elevates it above average difficulty. |
| Spec | 1.02y Partial fractions: decompose rational functions1.08e Area between curve and x-axis: using definite integrals1.08h Integration by substitution1.08j Integration using partial fractions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{3y-4}{y(3y+2)} \equiv \frac{A}{y} + \frac{B}{(3y+2)}\) | M1 | Writing in partial fractions and complete method for finding at least one of \(A\) or \(B\) |
| \(y=0 \Rightarrow -4 = 2A \Rightarrow A = -2\) | A1 | At least one of their \(A=-2\) or their \(B=9\) |
| \(y = -\frac{2}{3} \Rightarrow -6 = -\frac{2}{3}B \Rightarrow B=9\) | A1 | Both \(A=-2\) and \(B=9\) |
| \(\int \frac{3y-4}{y(3y+2)}\,dy = \int \frac{-2}{y} + \frac{9}{(3y+2)}\,dy\) | M1 | Integrates to give at least one of \(\frac{A}{y} \to \pm\lambda\ln y\) or \(\frac{B}{(3y+2)} \to \pm\mu\ln(3y+2)\), \(A\neq0\), \(B\neq0\) |
| \(= -2\ln y + 3\ln(3y+2) \{+c\}\) | A1 ft | At least one term correctly followed through from their \(A\) or \(B\) |
| \(-2\ln y + 3\ln(3y+2)\) or \(-2\ln y + 3\ln\!\left(y+\frac{2}{3}\right)\) | A1 cao | With correct bracketing, simplified or un-simplified. Can apply isw. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x = 4\sin^2\theta \Rightarrow \frac{dx}{d\theta} = 8\sin\theta\cos\theta\) or \(\frac{dx}{d\theta} = 4\sin 2\theta\) or \(dx = 8\sin\theta\cos\theta\,d\theta\) | B1 | |
| \(\int\sqrt{\frac{4\sin^2\theta}{4-4\sin^2\theta}}\cdot 8\sin\theta\cos\theta\,\{d\theta\}\) or \(\int\sqrt{\frac{4\sin^2\theta}{4-4\sin^2\theta}}\cdot 4\sin 2\theta\,\{d\theta\}\) | M1 | |
| \(= \int \underline{\tan\theta}\cdot 8\sin\theta\cos\theta\,\{d\theta\}\) or \(\int\underline{\tan\theta}\cdot 4\sin 2\theta\,\{d\theta\}\) | M1 | \(\sqrt{\frac{x}{4-x}} \to \pm K\tan\theta\) or \(\pm K\!\left(\frac{\sin\theta}{\cos\theta}\right)\) |
| \(= \int 8\sin^2\theta\,d\theta\) | A1 | \(\int 8\sin^2\theta\,d\theta\) including \(d\theta\) |
| \(3 = 4\sin^2\theta\) or \(\frac{3}{4} = \sin^2\theta\) or \(\sin\theta = \frac{\sqrt{3}}{2} \Rightarrow \theta = \frac{\pi}{3}\); \(x=0 \to \theta=0\) | B1 | Correct equation involving \(x=3\) leading to \(\theta=\frac{\pi}{3}\) and no incorrect work seen regarding limits |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(= \{8\}\int\left(\frac{1-\cos 2\theta}{2}\right)d\theta \left\{= \int(4-4\cos 2\theta)\,d\theta\right\}\) | M1 | Applies \(\cos 2\theta = 1 - 2\sin^2\theta\) to their integral |
| \(= \{8\}\!\left(\frac{1}{2}\theta - \frac{1}{4}\sin 2\theta\right) \left\{= 4\theta - 2\sin 2\theta\right\}\) | M1 | For \(\pm\alpha\theta \pm \beta\sin 2\theta\), \(\alpha,\beta\neq 0\) |
| \(\sin^2\theta \to \left(\frac{1}{2}\theta - \frac{1}{4}\sin 2\theta\right)\) | A1 | |
| \(\left\{\int_0^{\pi/3} 8\sin^2\theta\,d\theta\right\} = 8\!\left[\frac{1}{2}\theta - \frac{1}{4}\sin 2\theta\right]_0^{\pi/3} = 8\!\left(\left(\frac{\pi}{6} - \frac{1}{4}\cdot\frac{\sqrt{3}}{2}\right) - (0+0)\right)\) | ||
| \(= \frac{4}{3}\pi - \sqrt{3}\) | A1 o.e. | "Two term" exact answer e.g. \(\frac{4}{3}\pi - \sqrt{3}\) or \(\frac{1}{3}(4\pi - 3\sqrt{3})\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\int\frac{3y-4}{y(3y+2)}\,dy = \int\frac{6y+2}{3y^2+2y}\,dy - \int\frac{3y+6}{y(3y+2)}\,dy\) | ||
| \(\frac{3y+6}{y(3y+2)} \equiv \frac{A}{y}+\frac{B}{(3y+2)} \Rightarrow A=3, B=-6\) | M1 | See notes |
| At least one of \(A=3\) or \(B=-6\) | A1 | |
| Both \(A=3\) and \(B=-6\) | A1 | |
| Integrates to give at least one of \(\frac{M(6y+2)}{3y^2+2y} \to \pm\alpha\ln(3y^2+2y)\) or \(\frac{A}{y}\to\pm\lambda\ln y\) or \(\frac{B}{(3y+2)}\to\pm\mu\ln(3y+2)\), \(M\neq0\), \(A\neq0\), \(B\neq0\) | M1 | |
| At least one term correctly followed through | A1 ft | |
| \(= \ln(3y^2+2y) - 3\ln y + 2\ln(3y+2)\{+c\}\) | A1 cao | With correct bracketing, simplified or un-simplified |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{5}{y(3y+2)} \equiv \frac{A}{y}+\frac{B}{(3y+2)} \Rightarrow A=\frac{5}{2}, B=-\frac{15}{2}\) | M1 | See notes |
| At least one of \(A=\frac{5}{2}\) or \(B=-\frac{15}{2}\) | A1 | |
| Both \(A=\frac{5}{2}\) and \(B=-\frac{15}{2}\) | A1 | |
| Integrates to give \(\frac{M(3y+1)}{3y^2+2y}\to\pm\alpha\ln(3y^2+2y)\) or \(\frac{A}{y}\to\pm\lambda\ln y\) or \(\frac{B}{(3y+2)}\to\pm\mu\ln(3y+2)\) | M1 | |
| At least one term correctly followed through | A1 ft | |
| \(= \frac{1}{2}\ln(3y^2+2y) - \frac{5}{2}\ln y + \frac{5}{2}\ln(3y+2)\{+c\}\) | A1 cao | With correct bracketing, simplified or un-simplified |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{4}{y(3y+2)} \equiv \frac{A}{y}+\frac{B}{(3y+2)} \Rightarrow A=2, B=-6\) | M1 | See notes |
| At least one of \(A=2\) or \(B=-6\) | A1 | |
| Both \(A=2\) and \(B=-6\) | A1 | |
| Integrates to give \(\frac{C}{(3y+2)}\to\pm\alpha\ln(3y+2)\) or \(\frac{A}{y}\to\pm\lambda\ln y\) or \(\frac{B}{(3y+2)}\to\pm\mu\ln(3y+2)\), \(A\neq0\), \(B\neq0\), \(C\neq0\) | M1 | |
| At least one term correctly followed through | A1 ft | |
| \(= \ln(3y+2) - 2\ln y + 2\ln(3y+2)\{+c\}\) | A1 cao | With correct bracketing, simplified or un-simplified |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x=4\sin^2\theta \Rightarrow \frac{dx}{d\theta} = 8\sin\theta\cos\theta\) | B1 | As in Way 1 |
| \(\int\sqrt{\frac{4\sin^2\theta}{4-4\sin^2\theta}}\cdot 8\sin\theta\cos\theta\,\{d\theta\}\) | M1 | As before |
| \(= \int\frac{\sin\theta}{\sqrt{1-\sin^2\theta}}\cdot 8\cos\theta\sin\theta\,\{d\theta\}\) | ||
| \(= \int\sin\theta\cdot 8\sin\theta\,\{d\theta\}\) | M1 | Correct method leading to \(\sqrt{1-\sin^2\theta}\) being cancelled out |
| \(= \int 8\sin^2\theta\,d\theta\) | A1 cso | \(\int 8\sin^2\theta\,d\theta\) including \(d\theta\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x=4\sin^2\theta \Rightarrow \frac{dx}{d\theta} = 4\sin 2\theta\) | B1 | As in Way 1 |
| \(x = 4\sin^2\theta = 2-2\cos 2\theta\), \(4-x = 2+2\cos 2\theta\) | ||
| \(\int\sqrt{\frac{2-2\cos 2\theta}{2+2\cos 2\theta}}\cdot 4\sin 2\theta\,\{d\theta\}\) | M1 | |
| \(= \int\frac{2-2\cos 2\theta}{\sqrt{4-4\cos^2 2\theta}}\cdot 4\sin 2\theta\,\{d\theta\}\) | ||
| \(= \int\frac{2-2\cos 2\theta}{2\sin 2\theta}\cdot 4\sin 2\theta\,\{d\theta\} = \int 2(2-2\cos 2\theta)\,\{d\theta\}\) | M1 | Correct method leading to \(\sin 2\theta\) being cancelled out |
| \(= \int 8\sin^2\theta\,d\theta\) | A1 cso | \(\int 8\sin^2\theta\,d\theta\) including \(d\theta\) |
# Question 6:
## Part (i) — Way 1
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{3y-4}{y(3y+2)} \equiv \frac{A}{y} + \frac{B}{(3y+2)}$ | M1 | Writing in partial fractions and complete method for finding at least one of $A$ or $B$ |
| $y=0 \Rightarrow -4 = 2A \Rightarrow A = -2$ | A1 | At least one of their $A=-2$ **or** their $B=9$ |
| $y = -\frac{2}{3} \Rightarrow -6 = -\frac{2}{3}B \Rightarrow B=9$ | A1 | Both $A=-2$ **and** $B=9$ |
| $\int \frac{3y-4}{y(3y+2)}\,dy = \int \frac{-2}{y} + \frac{9}{(3y+2)}\,dy$ | M1 | Integrates to give at least one of $\frac{A}{y} \to \pm\lambda\ln y$ **or** $\frac{B}{(3y+2)} \to \pm\mu\ln(3y+2)$, $A\neq0$, $B\neq0$ |
| $= -2\ln y + 3\ln(3y+2) \{+c\}$ | A1 ft | At least one term correctly followed through from their $A$ or $B$ |
| $-2\ln y + 3\ln(3y+2)$ or $-2\ln y + 3\ln\!\left(y+\frac{2}{3}\right)$ | A1 cao | With correct bracketing, simplified or un-simplified. Can apply isw. |
**Total: [6]**
## Part (ii)(a) — Way 1
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x = 4\sin^2\theta \Rightarrow \frac{dx}{d\theta} = 8\sin\theta\cos\theta$ or $\frac{dx}{d\theta} = 4\sin 2\theta$ or $dx = 8\sin\theta\cos\theta\,d\theta$ | B1 | |
| $\int\sqrt{\frac{4\sin^2\theta}{4-4\sin^2\theta}}\cdot 8\sin\theta\cos\theta\,\{d\theta\}$ or $\int\sqrt{\frac{4\sin^2\theta}{4-4\sin^2\theta}}\cdot 4\sin 2\theta\,\{d\theta\}$ | M1 | |
| $= \int \underline{\tan\theta}\cdot 8\sin\theta\cos\theta\,\{d\theta\}$ or $\int\underline{\tan\theta}\cdot 4\sin 2\theta\,\{d\theta\}$ | M1 | $\sqrt{\frac{x}{4-x}} \to \pm K\tan\theta$ or $\pm K\!\left(\frac{\sin\theta}{\cos\theta}\right)$ |
| $= \int 8\sin^2\theta\,d\theta$ | A1 | $\int 8\sin^2\theta\,d\theta$ including $d\theta$ |
| $3 = 4\sin^2\theta$ or $\frac{3}{4} = \sin^2\theta$ or $\sin\theta = \frac{\sqrt{3}}{2} \Rightarrow \theta = \frac{\pi}{3}$; $x=0 \to \theta=0$ | B1 | Correct equation involving $x=3$ leading to $\theta=\frac{\pi}{3}$ and no incorrect work seen regarding limits |
**Total: [5]**
## Part (ii)(b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $= \{8\}\int\left(\frac{1-\cos 2\theta}{2}\right)d\theta \left\{= \int(4-4\cos 2\theta)\,d\theta\right\}$ | M1 | Applies $\cos 2\theta = 1 - 2\sin^2\theta$ to their integral |
| $= \{8\}\!\left(\frac{1}{2}\theta - \frac{1}{4}\sin 2\theta\right) \left\{= 4\theta - 2\sin 2\theta\right\}$ | M1 | For $\pm\alpha\theta \pm \beta\sin 2\theta$, $\alpha,\beta\neq 0$ |
| $\sin^2\theta \to \left(\frac{1}{2}\theta - \frac{1}{4}\sin 2\theta\right)$ | A1 | |
| $\left\{\int_0^{\pi/3} 8\sin^2\theta\,d\theta\right\} = 8\!\left[\frac{1}{2}\theta - \frac{1}{4}\sin 2\theta\right]_0^{\pi/3} = 8\!\left(\left(\frac{\pi}{6} - \frac{1}{4}\cdot\frac{\sqrt{3}}{2}\right) - (0+0)\right)$ | | |
| $= \frac{4}{3}\pi - \sqrt{3}$ | A1 o.e. | "Two term" exact answer e.g. $\frac{4}{3}\pi - \sqrt{3}$ or $\frac{1}{3}(4\pi - 3\sqrt{3})$ |
**Total: [4]**
---
## Part (i) — Way 2
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int\frac{3y-4}{y(3y+2)}\,dy = \int\frac{6y+2}{3y^2+2y}\,dy - \int\frac{3y+6}{y(3y+2)}\,dy$ | | |
| $\frac{3y+6}{y(3y+2)} \equiv \frac{A}{y}+\frac{B}{(3y+2)} \Rightarrow A=3, B=-6$ | M1 | See notes |
| At least one of $A=3$ **or** $B=-6$ | A1 | |
| Both $A=3$ **and** $B=-6$ | A1 | |
| Integrates to give at least one of $\frac{M(6y+2)}{3y^2+2y} \to \pm\alpha\ln(3y^2+2y)$ or $\frac{A}{y}\to\pm\lambda\ln y$ or $\frac{B}{(3y+2)}\to\pm\mu\ln(3y+2)$, $M\neq0$, $A\neq0$, $B\neq0$ | M1 | |
| At least one term correctly followed through | A1 ft | |
| $= \ln(3y^2+2y) - 3\ln y + 2\ln(3y+2)\{+c\}$ | A1 cao | With correct bracketing, simplified or un-simplified |
**Total: [6]**
## Part (i) — Way 3
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{5}{y(3y+2)} \equiv \frac{A}{y}+\frac{B}{(3y+2)} \Rightarrow A=\frac{5}{2}, B=-\frac{15}{2}$ | M1 | See notes |
| At least one of $A=\frac{5}{2}$ **or** $B=-\frac{15}{2}$ | A1 | |
| Both $A=\frac{5}{2}$ **and** $B=-\frac{15}{2}$ | A1 | |
| Integrates to give $\frac{M(3y+1)}{3y^2+2y}\to\pm\alpha\ln(3y^2+2y)$ or $\frac{A}{y}\to\pm\lambda\ln y$ or $\frac{B}{(3y+2)}\to\pm\mu\ln(3y+2)$ | M1 | |
| At least one term correctly followed through | A1 ft | |
| $= \frac{1}{2}\ln(3y^2+2y) - \frac{5}{2}\ln y + \frac{5}{2}\ln(3y+2)\{+c\}$ | A1 cao | With correct bracketing, simplified or un-simplified |
**Total: [6]**
## Part (i) — Way 4
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{4}{y(3y+2)} \equiv \frac{A}{y}+\frac{B}{(3y+2)} \Rightarrow A=2, B=-6$ | M1 | See notes |
| At least one of $A=2$ **or** $B=-6$ | A1 | |
| Both $A=2$ **and** $B=-6$ | A1 | |
| Integrates to give $\frac{C}{(3y+2)}\to\pm\alpha\ln(3y+2)$ or $\frac{A}{y}\to\pm\lambda\ln y$ or $\frac{B}{(3y+2)}\to\pm\mu\ln(3y+2)$, $A\neq0$, $B\neq0$, $C\neq0$ | M1 | |
| At least one term correctly followed through | A1 ft | |
| $= \ln(3y+2) - 2\ln y + 2\ln(3y+2)\{+c\}$ | A1 cao | With correct bracketing, simplified or un-simplified |
**Total: [6]**
## Part (ii)(a) — Way 2
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x=4\sin^2\theta \Rightarrow \frac{dx}{d\theta} = 8\sin\theta\cos\theta$ | B1 | As in Way 1 |
| $\int\sqrt{\frac{4\sin^2\theta}{4-4\sin^2\theta}}\cdot 8\sin\theta\cos\theta\,\{d\theta\}$ | M1 | As before |
| $= \int\frac{\sin\theta}{\sqrt{1-\sin^2\theta}}\cdot 8\cos\theta\sin\theta\,\{d\theta\}$ | | |
| $= \int\sin\theta\cdot 8\sin\theta\,\{d\theta\}$ | M1 | Correct method leading to $\sqrt{1-\sin^2\theta}$ being cancelled out |
| $= \int 8\sin^2\theta\,d\theta$ | A1 cso | $\int 8\sin^2\theta\,d\theta$ including $d\theta$ |
## Part (ii)(a) — Way 3
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x=4\sin^2\theta \Rightarrow \frac{dx}{d\theta} = 4\sin 2\theta$ | B1 | As in Way 1 |
| $x = 4\sin^2\theta = 2-2\cos 2\theta$, $4-x = 2+2\cos 2\theta$ | | |
| $\int\sqrt{\frac{2-2\cos 2\theta}{2+2\cos 2\theta}}\cdot 4\sin 2\theta\,\{d\theta\}$ | M1 | |
| $= \int\frac{2-2\cos 2\theta}{\sqrt{4-4\cos^2 2\theta}}\cdot 4\sin 2\theta\,\{d\theta\}$ | | |
| $= \int\frac{2-2\cos 2\theta}{2\sin 2\theta}\cdot 4\sin 2\theta\,\{d\theta\} = \int 2(2-2\cos 2\theta)\,\{d\theta\}$ | M1 | Correct method leading to $\sin 2\theta$ being cancelled out |
| $= \int 8\sin^2\theta\,d\theta$ | A1 cso | $\int 8\sin^2\theta\,d\theta$ including $d\theta$ |6. (i) Given that $y > 0$, find
$$\int \frac { 3 y - 4 } { y ( 3 y + 2 ) } d y$$
(ii) (a) Use the substitution $x = 4 \sin ^ { 2 } \theta$ to show that
$$\int _ { 0 } ^ { 3 } \sqrt { \left( \frac { x } { 4 - x } \right) } \mathrm { d } x = \lambda \int _ { 0 } ^ { \frac { \pi } { 3 } } \sin ^ { 2 } \theta \mathrm {~d} \theta$$
where $\lambda$ is a constant to be determined.\\
(b) Hence use integration to find
$$\int _ { 0 } ^ { 3 } \sqrt { \left( \frac { x } { 4 - x } \right) } d x$$
giving your answer in the form $a \pi + b$, where $a$ and $b$ are exact constants.
\hfill \mbox{\textit{Edexcel C4 2016 Q6 [15]}}