| Exam Board | OCR |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2006 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration using inverse trig and hyperbolic functions |
| Type | Half-angle tangent substitution t = tan(x/2) |
| Difficulty | Challenging +1.2 This is a structured Further Maths question with clear guidance through both parts. Part (i) is routine completing the square, and part (ii) applies the standard Weierstrass substitution with explicit instruction. While the algebraic manipulation requires care and the final arctan evaluation needs precision, the method is prescribed and follows a well-practiced technique for FP2 students. It's moderately harder than average A-level due to being Further Maths content, but straightforward within that context. |
| Spec | 1.02e Complete the square: quadratic polynomials and turning points1.08h Integration by substitution |
| Answer | Marks | Guidance |
|---|---|---|
| (i) Get \((t + \frac{1}{2})^2 + \frac{3}{4}\) | B1 | cao |
| (ii) Derive or quote \(\frac{dx}{dt} = \frac{2}{1+t^2}\) | B1 | |
| Derive or quote \(\sin x = \frac{2t}{(1 + t^2)}\) | B1 | |
| Attempt to replace all \(x\) and \(dx\) | M1 | |
| Get integral of form \(\frac{A}{(Bt^2+C | +D)}\) | A1 |
| Use complete square form as \(\tan^{-1}(f(t))\) | M1 | From formulae book or substitution |
| Get A.G. | A1 |
**(i)** Get $(t + \frac{1}{2})^2 + \frac{3}{4}$ | B1 | cao
**(ii)** Derive or quote $\frac{dx}{dt} = \frac{2}{1+t^2}$ | B1 |
Derive or quote $\sin x = \frac{2t}{(1 + t^2)}$ | B1 |
Attempt to replace all $x$ and $dx$ | M1 |
Get integral of form $\frac{A}{(Bt^2+C|+D)}$ | A1 | From their expressions, $C \neq 0$
Use complete square form as $\tan^{-1}(f(t))$ | M1 | From formulae book or substitution
Get A.G. | A1 |5 (i) Express $t ^ { 2 } + t + 1$ in the form $( t + a ) ^ { 2 } + b$.\\
(ii) By using the substitution $\tan \frac { 1 } { 2 } x = t$, show that
$$\int _ { 0 } ^ { \frac { 1 } { 2 } \pi } \frac { 1 } { 2 + \sin x } \mathrm {~d} x = \frac { \sqrt { 3 } } { 9 } \pi$$
\hfill \mbox{\textit{OCR FP2 2006 Q5 [7]}}