| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9794/2 (Pre-U Mathematics Paper 2) |
| Session | Specimen |
| Marks | 6 |
| Topic | Standard trigonometric equations |
| Type | Prove identity then solve |
| Difficulty | Standard +0.8 This question requires knowledge of half-angle substitutions (t-formulae), which are beyond standard A-level content and typically appear in Further Maths or Pre-U. Part (i) involves deriving standard identities using double-angle formulae, while part (ii) applies these to evaluate a definite integral—a multi-step problem requiring both algebraic manipulation and integration technique. However, once the substitution framework is understood, the execution is relatively straightforward, making it moderately challenging rather than extremely difficult. |
| Spec | 4.08h Integration: inverse trig/hyperbolic substitutions |
**(i)** Using $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A\tan B}$, for $A = B = \frac{1}{2}x$: B1
Then $\tan x = \frac{2t}{1-t^2}$ **AG**
With the restriction on $t$, a right angled triangle exists with M1
Opposite $= 2t$, adjacent $= 1-t^2$ and hypotenuse $= 1+t^2$
$\Rightarrow \sin x = \frac{2t}{1+t^2}$ for $0 \leq t < 1$ A1 **[3]**
**(ii)** $dt = \frac{1}{2}\sec^2\frac{1}{2}x\,dx \Rightarrow dx = \frac{2}{1+t^2}\,dt$ B1
The new limits are $t = 0$ and $t = \frac{1}{\sqrt{3}}$ B1
On correct substitution, the integral becomes: M1
$\int_0^{1/\sqrt{3}} \frac{1}{1 + \frac{2t}{1+t^2}} \times \frac{2}{1+t^2}\,dt = \int_0^{1/\sqrt{3}} \frac{2}{(1+t)^2}\,dt$ A1
Correct substitution of the limits: M1
$= \left[\frac{-2}{1+t}\right]_0^{1/\sqrt{3}} = \frac{-2\sqrt{3}}{\sqrt{3}+1} + 2 = \frac{2}{\sqrt{3}+1} = \frac{2}{\sqrt{3}+1} \times \frac{\sqrt{3}-1}{\sqrt{3}-1} = \sqrt{3}-1$ **AG** A1 **[6]**8 (i) Show that
$$\tan x = \frac { 2 t } { 1 - t ^ { 2 } } \text { for } 0 \leq t < 1 , \text { where } t = \tan \frac { 1 } { 2 } x$$
and deduce that
$$\sin x = \frac { 2 t } { 1 + t ^ { 2 } }$$
(ii) Using the substitution $t = \tan \frac { 1 } { 2 } x$, show that
$$\int _ { 0 } ^ { \frac { 1 } { 3 } \pi } \frac { 1 } { 1 + \sin x } \mathrm {~d} x = \sqrt { 3 } - 1$$
\hfill \mbox{\textit{Pre-U Pre-U 9794/2 Q8 [6]}}