| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9794/2 (Pre-U Mathematics Paper 2) |
| Year | 2016 |
| Session | June |
| Marks | 11 |
| Topic | Reciprocal Trig & Identities |
| Type | Integration using reciprocal identities |
| Difficulty | Challenging +1.2 Part (i) requires algebraic manipulation of a trigonometric identity, working backwards from a non-obvious form involving sec and tan - this demands insight beyond routine identities. Part (ii) then requires integrating sec²x and sec x tan x (standard but not trivial) and careful arithmetic with the limits. The multi-step nature, non-standard identity form, and integration of sec functions place this above average difficulty but within reach of well-prepared students. |
| Spec | 1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.08c Integrate e^(kx), 1/x, sin(kx), cos(kx) |
| Answer | Marks | Guidance |
|---|---|---|
| (i) | B1 | \(\sec x = \frac{1}{\cos x}\) oe seen anywhere. |
| \(\frac{\sin x}{1+\sin x} = \frac{\sin x(1-\sin x)}{(1+\sin x)(1-\sin x)} = \frac{\sin x - \sin^2 x}{1-\sin^2 x}\) | M1, A1 | Multiply top and bottom by \(1-\sin x\). Obtain correct unsimplified expression. |
| \(= \frac{\sin x - 1 + \cos^2 x}{\cos^2 x}\) | M1 | Write denominator as \(\cos^2 x\). |
| \(= \sec x \tan x - \sec^2 x + 1\) | A1 [5] | Obtain correct simplified expression. |
| Answer | Marks | Guidance |
|---|---|---|
| \(\sec x \tan x - \sec^2 x + 1 \equiv \frac{\sin x - 1 + \cos^2 x}{\cos^2 x} = \frac{\sin x - \sin^2 x}{1 - \sin^2 x}\) | M1, M1, A1, A1 | Write with common denominator of \(\cos^2 x\). Attempt expression in terms of \(\sin x\) only. Obtain correct unsimplified expression. Obtain correct simplified expression. |
| (ii) \(\int_0^{\pi/4} \frac{\sin x}{1+\sin x} dx = \int_0^{\pi/4} \sec x \tan x - \sec^2 x + 1 dx = [\sec x - \tan x + x]_0^{\pi/4}\) | M1, A1, A1 | Attempt integration of given expression (at least two terms). Obtain at least two correct terms (allow if third term not yet integrated). Obtain fully correct integral. |
| \(= (\sqrt{2} - 1 + \frac{1}{4}\pi) - (1 - 0 + 0)\) | M1 | Attempt correct use of limits (correct order and subtraction) in their integration attempt. |
| \(= \frac{1}{4}\pi + \sqrt{2} - 2\) AG | B1, A1 [6] | State or imply \(\sec \frac{1}{4}\pi = \sqrt{2}\). Obtain given answer convincingly. |
**(i)** | B1 | $\sec x = \frac{1}{\cos x}$ oe seen anywhere.
$\frac{\sin x}{1+\sin x} = \frac{\sin x(1-\sin x)}{(1+\sin x)(1-\sin x)} = \frac{\sin x - \sin^2 x}{1-\sin^2 x}$ | M1, A1 | Multiply top and bottom by $1-\sin x$. Obtain correct unsimplified expression.
$= \frac{\sin x - 1 + \cos^2 x}{\cos^2 x}$ | M1 | Write denominator as $\cos^2 x$.
$= \sec x \tan x - \sec^2 x + 1$ | A1 [5] | Obtain correct simplified expression.
**OR**
$\sec x \tan x - \sec^2 x + 1 \equiv \frac{\sin x - 1 + \cos^2 x}{\cos^2 x} = \frac{\sin x - \sin^2 x}{1 - \sin^2 x}$ | M1, M1, A1, A1 | Write with common denominator of $\cos^2 x$. Attempt expression in terms of $\sin x$ only. Obtain correct unsimplified expression. Obtain correct simplified expression.
**(ii)** $\int_0^{\pi/4} \frac{\sin x}{1+\sin x} dx = \int_0^{\pi/4} \sec x \tan x - \sec^2 x + 1 dx = [\sec x - \tan x + x]_0^{\pi/4}$ | M1, A1, A1 | Attempt integration of given expression (at least two terms). Obtain at least two correct terms (allow if third term not yet integrated). Obtain fully correct integral.
$= (\sqrt{2} - 1 + \frac{1}{4}\pi) - (1 - 0 + 0)$ | M1 | Attempt correct use of limits (correct order and subtraction) in their integration attempt.
$= \frac{1}{4}\pi + \sqrt{2} - 2$ **AG** | B1, A1 [6] | State or imply $\sec \frac{1}{4}\pi = \sqrt{2}$. Obtain given answer convincingly.
Allow non 'hence' methods.\begin{enumerate}[label=(\roman*)]
\item Show that $\frac{\sin x}{1 + \sin x} \equiv \sec x \tan x - \sec^2 x + 1$. [5]
\item Hence show that $\int_0^{\frac{\pi}{4}} \frac{\sin x}{1 + \sin x} \, dx = \frac{1}{4}\pi + \sqrt{2} - 2$. [6]
\end{enumerate}
\hfill \mbox{\textit{Pre-U Pre-U 9794/2 2016 Q9 [11]}}