| Exam Board | OCR |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Session | Specimen |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration with Partial Fractions |
| Type | Complex partial fractions with multiple techniques |
| Difficulty | Challenging +1.8 This FP2 question combines the Weierstrass t-substitution with partial fractions involving both linear and irreducible quadratic factors. Part (i) requires careful algebraic manipulation of trigonometric identities under the substitution, part (ii) needs handling of mixed partial fractions (linear + quadratic), and part (iii) involves integrating ln and arctan terms. While systematic, it demands multiple advanced techniques, extended algebraic stamina, and careful bookkeeping across three connected parts—significantly harder than typical A-level but standard for Further Maths FP2. |
| Spec | 1.02y Partial fractions: decompose rational functions1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.08d Evaluate definite integrals: between limits1.08h Integration by substitution |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(\frac{df}{dx} = \frac{1}{2}(1+t^2)\) | B1 | For this relation, stated or used |
| \(\int_0^{\frac{\pi}{4}} \frac{1-\cos x}{1+\sin x} \, dx = \int_0^1 \frac{1-\frac{1-t^2}{1+t^2}}{1+\frac{2t}{1+t^2}} \cdot \frac{2}{1+t^2} \, dt\) | M1 | For complete substitution for \(x\) in integrand |
| \(-\) | B1 | For justification of limits \(0\) and \(1\) for \(t\) |
| \(= 2\sqrt{2}\int_0^1 \frac{2t^2}{(1+t)^2(1+t^2)} - \frac{2}{1+t^2} \, dt = 2\sqrt{2}\int_0^1 \frac{t}{(1+t)(1+t^2)} \, dt\) | A1 | For correct simplification to given answer |
| \(-\) | 4 | |
| (ii) \(\frac{t}{(1+t)(1+t^2)} = \frac{A}{1+t} + \frac{Bt+C}{1+t^2}\) | B1 | For statement of correct form of pfs |
| Hence \(t = A(1+t^2) + (Bt+C)(1+t)\) | M1 | For any use of the identity involving \(B\) or \(C\) |
| From which \(A = -\frac{1}{2}\), \(B = \frac{1}{2}\), \(C = \frac{1}{2}\) | B1 | For correct value of \(A\) |
| \(-\) | A1 | For correct value of \(B\) |
| \(-\) | A1 | For correct value of \(C\) |
| \(-\) | 5 | |
| (iii) \(\operatorname{Int} = 2\sqrt{2}\left[-\frac{1}{2}\ln(1+t) + \frac{1}{4}\ln(1+t^2) + \frac{1}{2}\tan^{-1}t\right]_0^1\) | B1√ | For both logarithm terms correct |
| \(-\) | B1√ | For the inverse tan term correct |
| \(= \frac{1}{4}(\pi - 2\ln 2)\sqrt{2}\) | M1 | For use of appropriate limits |
| \(-\) | A1 | For correct (exact) answer in any form |
| \(-\) | 4 |
(i) $\frac{df}{dx} = \frac{1}{2}(1+t^2)$ | B1 | For this relation, stated or used |
$\int_0^{\frac{\pi}{4}} \frac{1-\cos x}{1+\sin x} \, dx = \int_0^1 \frac{1-\frac{1-t^2}{1+t^2}}{1+\frac{2t}{1+t^2}} \cdot \frac{2}{1+t^2} \, dt$ | M1 | For complete substitution for $x$ in integrand |
$-$ | B1 | For justification of limits $0$ and $1$ for $t$ |
$= 2\sqrt{2}\int_0^1 \frac{2t^2}{(1+t)^2(1+t^2)} - \frac{2}{1+t^2} \, dt = 2\sqrt{2}\int_0^1 \frac{t}{(1+t)(1+t^2)} \, dt$ | A1 | For correct simplification to given answer |
$-$ | **4** | |
(ii) $\frac{t}{(1+t)(1+t^2)} = \frac{A}{1+t} + \frac{Bt+C}{1+t^2}$ | B1 | For statement of correct form of pfs |
Hence $t = A(1+t^2) + (Bt+C)(1+t)$ | M1 | For any use of the identity involving $B$ or $C$ |
From which $A = -\frac{1}{2}$, $B = \frac{1}{2}$, $C = \frac{1}{2}$ | B1 | For correct value of $A$ |
$-$ | A1 | For correct value of $B$ |
$-$ | A1 | For correct value of $C$ |
$-$ | **5** | |
(iii) $\operatorname{Int} = 2\sqrt{2}\left[-\frac{1}{2}\ln(1+t) + \frac{1}{4}\ln(1+t^2) + \frac{1}{2}\tan^{-1}t\right]_0^1$ | B1√ | For both logarithm terms correct |
$-$ | B1√ | For the inverse tan term correct |
$= \frac{1}{4}(\pi - 2\ln 2)\sqrt{2}$ | M1 | For use of appropriate limits |
$-$ | A1 | For correct (exact) answer in any form |
$-$ | **4** | |8 (i) Use the substitution $t = \tan \frac { 1 } { 2 } x$ to show that
$$\int _ { 0 } ^ { \frac { 1 } { 2 } \pi } \sqrt { \frac { 1 - \cos x } { 1 + \sin x } } \mathrm {~d} x = 2 \sqrt { } 2 \int _ { 0 } ^ { 1 } \frac { t } { ( 1 + t ) \left( 1 + t ^ { 2 } \right) } \mathrm { d } t$$
(ii) Express $\frac { t } { ( 1 + t ) \left( 1 + t ^ { 2 } \right) }$ in partial fractions.\\
(iii) Hence find $\int _ { 0 } ^ { \frac { 1 } { 2 } \pi } \sqrt { \frac { 1 - \cos x } { 1 + \sin x } } \mathrm {~d} x$, expressing your answer in an exact form.
\hfill \mbox{\textit{OCR FP2 Q8 [13]}}