| Exam Board | OCR |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2009 |
| Session | June |
| Marks | 7 |
| 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 Further Maths FP2 question requiring the half-angle (Weierstrass) substitution t = tan(θ/2), which is a standard but non-trivial technique. Part (i) requires careful manipulation of trigonometric identities and changing limits, while part (ii) is straightforward integration. The question tests technical proficiency with a specialized substitution method beyond standard A-level, placing it moderately above average difficulty. |
| Spec | 1.05k Further identities: sec^2=1+tan^2 and cosec^2=1+cot^21.08h Integration by substitution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Derive/quote \(d\theta = \frac{2dt}{1+t^2}\) | B1 | May be implied |
| Replace their \(\cos\theta\) and their \(d\theta\), both in terms of \(t\) | M1 | Not \(d\theta = dt\) |
| Clearly get \(\int \frac{(1-t^2)}{(1+t^2)} dt\) or equiv | A1 | Accept limits of \(t\) quoted here |
| Attempt to divide out | M1 | Or use AG to get answer above |
| Clearly get/derive AG | A1 | |
| SC: Derive \(d\theta = 2\cos^2\frac{1}{2}\theta\, dt\) | B1 | |
| Replace \(\cos\theta\) in terms of half-angles and their \(d\theta\) \((\neq dt)\) | M1 | |
| Get \(\int 2\cos^2\frac{1}{2}\theta - 1\, dt\) or \(\int 1 - \frac{1}{2}\cos^2\frac{1}{2}\theta \cdot \frac{2}{1+t^2}\, dt\) | A1 | |
| Use \(\sec^2\frac{1}{2}\theta = 1+t^2\) | M1 | |
| Clearly get/derive AG | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Integrate to \(a\tan^{-1}bt - t\) | M1 | |
| Get \(\frac{1}{2}\pi - 1\) | A1 |
## Question 5(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Derive/quote $d\theta = \frac{2dt}{1+t^2}$ | B1 | May be implied |
| Replace their $\cos\theta$ and their $d\theta$, both in terms of $t$ | M1 | Not $d\theta = dt$ |
| Clearly get $\int \frac{(1-t^2)}{(1+t^2)} dt$ or equiv | A1 | Accept limits of $t$ quoted here |
| Attempt to divide out | M1 | Or use AG to get answer above |
| Clearly get/derive AG | A1 | |
| **SC:** Derive $d\theta = 2\cos^2\frac{1}{2}\theta\, dt$ | B1 | |
| Replace $\cos\theta$ in terms of half-angles and their $d\theta$ $(\neq dt)$ | M1 | |
| Get $\int 2\cos^2\frac{1}{2}\theta - 1\, dt$ or $\int 1 - \frac{1}{2}\cos^2\frac{1}{2}\theta \cdot \frac{2}{1+t^2}\, dt$ | A1 | |
| Use $\sec^2\frac{1}{2}\theta = 1+t^2$ | M1 | |
| Clearly get/derive AG | A1 | |
---
## Question 5(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Integrate to $a\tan^{-1}bt - t$ | M1 | |
| Get $\frac{1}{2}\pi - 1$ | A1 | |
---5 It is given that $I = \int _ { 0 } ^ { \frac { 1 } { 2 } \pi } \frac { \cos \theta } { 1 + \cos \theta } \mathrm { d } \theta$.\\
(i) By using the substitution $t = \tan \frac { 1 } { 2 } \theta$, show that $I = \int _ { 0 } ^ { 1 } \left( \frac { 2 } { 1 + t ^ { 2 } } - 1 \right) \mathrm { d } t$.\\
(ii) Hence find $I$ in terms of $\pi$.
\hfill \mbox{\textit{OCR FP2 2009 Q5 [7]}}