| Exam Board | OCR |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2006 |
| Session | January |
| Marks | 9 |
| 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 question requires students to verify a non-trivial trigonometric substitution involving sin²θ (not just sinθ), correctly derive dx = 2sinθcosθ dθ, simplify the resulting expression using Pythagorean identities, then apply the double angle formula to integrate sin²θ and evaluate definite limits. While the substitution is given, the algebraic manipulation and multi-step process elevates this above a standard C4 integration question. |
| Spec | 1.08h Integration by substitution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Attempt to connect \(dx\), \(d\theta\) | M1 | But not \(dx = d\theta\) |
| \(dx = 2\sin\theta\cos\theta\, d\theta\) | A1 | AEF |
| \(\sqrt{\frac{x}{1-x}} = \frac{\sin\theta}{\cos\theta}\) | B1 | Ignore any references to \(\pm\) |
| Reduction to \(\int 2\sin^2\theta\, d\theta\) | A1 | 4 AG WWW |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\sin^2\theta = k(+/-1+/-\cos 2\theta)\) | M1 | Attempt to change \((2)\sin^2\theta\) into \(f(\cos 2\theta)\) |
| \(2\sin^2\theta = 1-\cos 2\theta\) | A1 | Correct attempt |
| \(\int\cos 2\theta\, d\theta = \frac{1}{2}\sin 2\theta\) | B1 | Seen anywhere in this part |
| Attempting to change limits | M1 | Or attempting to resubstitute; Accept degrees |
| \(\frac{1}{2}\pi\) | A1 | 5 |
| Alternatively: Parts once & use \(\cos^2\theta = 1-\sin^2\theta\) | (M2) | Instead of the M1 A1 B1 |
| \(\frac{1}{2}(\theta - \sin\theta\cos\theta)\) | (A1) | Then the final M1 A1 for use of limits |
# Question 6:
## Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Attempt to connect $dx$, $d\theta$ | M1 | But not $dx = d\theta$ |
| $dx = 2\sin\theta\cos\theta\, d\theta$ | A1 | AEF |
| $\sqrt{\frac{x}{1-x}} = \frac{\sin\theta}{\cos\theta}$ | B1 | Ignore any references to $\pm$ |
| Reduction to $\int 2\sin^2\theta\, d\theta$ | A1 | **4** AG WWW |
## Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\sin^2\theta = k(+/-1+/-\cos 2\theta)$ | M1 | Attempt to change $(2)\sin^2\theta$ into $f(\cos 2\theta)$ |
| $2\sin^2\theta = 1-\cos 2\theta$ | A1 | Correct attempt |
| $\int\cos 2\theta\, d\theta = \frac{1}{2}\sin 2\theta$ | B1 | Seen anywhere in this part |
| Attempting to change limits | M1 | Or attempting to resubstitute; Accept degrees |
| $\frac{1}{2}\pi$ | A1 | **5** |
| Alternatively: Parts once & use $\cos^2\theta = 1-\sin^2\theta$ | (M2) | Instead of the M1 A1 B1 |
| $\frac{1}{2}(\theta - \sin\theta\cos\theta)$ | (A1) | Then the final M1 A1 for use of limits |6 (i) Show that the substitution $x = \sin ^ { 2 } \theta$ transforms $\int \sqrt { \frac { x } { 1 - x } } \mathrm {~d} x$ to $\int 2 \sin ^ { 2 } \theta \mathrm {~d} \theta$.\\
(ii) Hence find $\int _ { 0 } ^ { 1 } \sqrt { \frac { x } { 1 - x } } \mathrm {~d} x$.
\hfill \mbox{\textit{OCR C4 2006 Q6 [9]}}