| Exam Board | Edexcel |
|---|---|
| Module | Paper 1 (Paper 1) |
| Session | Specimen |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration by Substitution |
| Type | Show definite integral equals specific value (trigonometric substitution or identity) |
| Difficulty | Standard +0.8 This is a 'show that' integration problem requiring substitution with trigonometric identities. Students must recognize to use sin 2θ = 2sin θ cos θ, simplify the integrand, apply an appropriate substitution (u = 1 + cos θ), and evaluate to match the given answer including a logarithmic term. It requires multiple techniques and careful algebraic manipulation, making it moderately challenging but still within standard A-level scope. |
| Spec | 1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.08h Integration by substitution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Applies substitution \(u = 1 + \cos\theta\); achieves \(\int \ldots\frac{(u-1)}{u}\ldots\) | M1 | AO 3.1a |
| \(u = 1 + \cos\theta \Rightarrow \frac{du}{d\theta} = -\sin\theta\) and \(\sin 2\theta = 2\sin\theta\cos\theta\) | M1 | AO 1.1b |
| \(\int \frac{\sin 2\theta}{1+\cos\theta}\,d\theta = \int \frac{2\sin\theta\cos\theta}{1+\cos\theta}\,d\theta = \int \frac{-2(u-1)}{u}\,du\) | A1 | AO 2.1 |
| \(-2\int\left(1 - \frac{1}{u}\right)du\) | M1 | AO 1.1b |
| \(= -2(u - \ln u)\) | M1 | AO 1.1b |
| \(\left[\int_0^{\frac{\pi}{2}} \frac{\sin 2\theta}{1+\cos\theta}\,d\theta\right] = -2\left[u - \ln u\right]_2^1 = -2((1-\ln 1)-(2-\ln 2))\) | M1 | AO 1.1b |
| \(= -2(-1 + \ln 2) = 2 - 2\ln 2\) | A1* | AO 2.1 — given answer, no errors seen |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Achieves \(\int\ldots\frac{u}{u+1}\ldots\) | M1 | AO 3.1a |
| \(u = \cos\theta \Rightarrow \frac{du}{d\theta} = -\sin\theta\) and \(\sin 2\theta = 2\sin\theta\cos\theta\) | M1 | AO 1.1b |
| \(\int \frac{\sin 2\theta}{1+\cos\theta}\,d\theta = \int \frac{-2u}{u+1}\,du\) | A1 | AO 2.1 |
| \(-2\int\frac{(u+1)-1}{u+1}\,du = -2\int\left(1 - \frac{1}{u+1}\right)du\) | M1 | AO 1.1b |
| \(= -2(u - \ln(u+1))\) | M1 | AO 1.1b |
| \(= -2\left[u - \ln(u+1)\right]_1^0 = -2((0-\ln 1)-(1-\ln 2))\) | M1 | AO 1.1b |
| \(= -2(-1+\ln 2) = 2 - 2\ln 2\) | A1* | AO 2.1 — given answer, no errors seen |
## Question 12:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Applies substitution $u = 1 + \cos\theta$; achieves $\int \ldots\frac{(u-1)}{u}\ldots$ | M1 | AO 3.1a |
| $u = 1 + \cos\theta \Rightarrow \frac{du}{d\theta} = -\sin\theta$ and $\sin 2\theta = 2\sin\theta\cos\theta$ | M1 | AO 1.1b |
| $\int \frac{\sin 2\theta}{1+\cos\theta}\,d\theta = \int \frac{2\sin\theta\cos\theta}{1+\cos\theta}\,d\theta = \int \frac{-2(u-1)}{u}\,du$ | A1 | AO 2.1 |
| $-2\int\left(1 - \frac{1}{u}\right)du$ | M1 | AO 1.1b |
| $= -2(u - \ln u)$ | M1 | AO 1.1b |
| $\left[\int_0^{\frac{\pi}{2}} \frac{\sin 2\theta}{1+\cos\theta}\,d\theta\right] = -2\left[u - \ln u\right]_2^1 = -2((1-\ln 1)-(2-\ln 2))$ | M1 | AO 1.1b |
| $= -2(-1 + \ln 2) = 2 - 2\ln 2$ | A1* | AO 2.1 — given answer, no errors seen |
**Alt 1** (substitution $u = \cos\theta$):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Achieves $\int\ldots\frac{u}{u+1}\ldots$ | M1 | AO 3.1a |
| $u = \cos\theta \Rightarrow \frac{du}{d\theta} = -\sin\theta$ and $\sin 2\theta = 2\sin\theta\cos\theta$ | M1 | AO 1.1b |
| $\int \frac{\sin 2\theta}{1+\cos\theta}\,d\theta = \int \frac{-2u}{u+1}\,du$ | A1 | AO 2.1 |
| $-2\int\frac{(u+1)-1}{u+1}\,du = -2\int\left(1 - \frac{1}{u+1}\right)du$ | M1 | AO 1.1b |
| $= -2(u - \ln(u+1))$ | M1 | AO 1.1b |
| $= -2\left[u - \ln(u+1)\right]_1^0 = -2((0-\ln 1)-(1-\ln 2))$ | M1 | AO 1.1b |
| $= -2(-1+\ln 2) = 2 - 2\ln 2$ | A1* | AO 2.1 — given answer, no errors seen |
**Total: 7 marks**
---\begin{enumerate}
\item Show that
\end{enumerate}
$$\int _ { 0 } ^ { \frac { \pi } { 2 } } \frac { \sin 2 \theta } { 1 + \cos \theta } d \theta = 2 - 2 \ln 2$$
\hfill \mbox{\textit{Edexcel Paper 1 Q12 [7]}}