Standard +0.3 This is a straightforward integration by substitution question with a clear hint in the integrand structure. The substitution u = 3 + cos(2x) is natural given the denominator, and sin(2x) dx conveniently provides the derivative needed. Evaluating limits and simplifying ln(4/3) - ln(3) = ln(4/3) requires only basic log laws. Slightly easier than average as the substitution is obvious and the algebra is clean.
\(\int -\frac{1}{2u}\,du\), or if \(v = \cos 2x\), \(\int -\frac{1}{2(3+v)}\,dv\)
\[= \left[-\frac{1}{2}\ln u\right]_4^2\]
A1
\(-\frac{1}{2}\ln u\,\Big]\) or \(\left[-\frac{1}{2}\ln(3+v)\right]\) ignore incorrect limits
\[= -\frac{1}{2}\ln 2 + \frac{1}{2}\ln 4\]
A1
from correct working o.e. e.g. \(-\frac{1}{2}\ln(3+\cos(2\cdot\pi/2)) + \frac{1}{2}\ln(3 + \cos(2\cdot 0))\); o.e. required step for final A1, must have evaluated to 4 and 2 at this stage
\[= \frac{1}{2}\ln\!\left(\frac{4}{2}\right)\]
Answer
Marks
Guidance
\[= \frac{1}{2}\ln 2 \text{ *}\]
A1
NB AG
[5]
## Question 5:
$$\int_0^{\pi/2} \frac{\sin 2x}{3 + \cos 2x}\,dx = \left[-\frac{1}{2}\ln(3 + \cos 2x)\right]_0^{\pi/2}$$ | M1 | $k\ln(3 + \cos 2x)$
| A2 | $-\frac{1}{2}\ln(3 + \cos 2x)$
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*or* $u = 3 + \cos 2x$, $du = -2\sin 2x\,dx$ | M1 | o.e. e.g. $du/dx = -2\sin 2x$ or if $v = \cos 2x$, $dv = -2\sin 2x\,dx$ o.e. condone $2\sin 2x\,dx$
$$\int_0^{\pi/2} \frac{\sin 2x}{3 + \cos 2x}\,dx = \int_4^2 -\frac{1}{2u}\,du$$ | A1 | $\int -\frac{1}{2u}\,du$, or if $v = \cos 2x$, $\int -\frac{1}{2(3+v)}\,dv$
$$= \left[-\frac{1}{2}\ln u\right]_4^2$$ | A1 | $-\frac{1}{2}\ln u\,\Big]$ or $\left[-\frac{1}{2}\ln(3+v)\right]$ ignore incorrect limits
$$= -\frac{1}{2}\ln 2 + \frac{1}{2}\ln 4$$ | A1 | from correct working o.e. e.g. $-\frac{1}{2}\ln(3+\cos(2\cdot\pi/2)) + \frac{1}{2}\ln(3 + \cos(2\cdot 0))$; o.e. required step for final A1, must have evaluated to 4 and 2 at this stage
$$= \frac{1}{2}\ln\!\left(\frac{4}{2}\right)$$
$$= \frac{1}{2}\ln 2 \text{ *}$$ | A1 | **NB AG**
**[5]**