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 strongly suggested by the denominator, and sin(2x) in the numerator is conveniently its derivative (up to a constant). After substitution, it reduces to a standard ln integral, followed by routine evaluation of limits. Slightly easier than average due to the obvious substitution and direct path to the answer.
\(\int -\frac{1}{2u}\,du\), or if \(v = \cos 2x\), \(\int -\frac{1}{2(3+v)}\,dv\)
\(= \left[-\frac{1}{2}\ln u\right]_4^2\)
A1
\([-\frac{1}{2}\ln u]\) or \([-\frac{1}{2}\ln(3+v)]\) ignore incorrect limits
\(= -\frac{1}{2}\ln 2 + \frac{1}{2}\ln 4\)
A1
from correct working o.e. e.g. \(-\frac{1}{2}\ln(3+\cos(2\cdot\frac{\pi}{2})) + \frac{1}{2}\ln(3+\cos(2\cdot 0))\); o.e. required step for final A1, must have evaluated to 4 and 2 at this stage
\(= \frac{1}{2}\ln\frac{4}{2}\)
\(= \frac{1}{2}\ln 2\) *
A1
NB AG
## Question 6:
| Answer | Mark | Guidance |
|--------|------|----------|
| $\int_0^{\pi/2} \frac{\sin 2x}{3+\cos 2x}\,dx = \left[-\frac{1}{2}\ln(3+\cos 2x)\right]_0^{\pi/2}$ | M1 | $k\ln(3+\cos 2x)$ |
| | A2 | $-\frac{1}{2}\ln(3+\cos 2x)$ |
| 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 | $[-\frac{1}{2}\ln u]$ or $[-\frac{1}{2}\ln(3+v)]$ ignore incorrect limits |
| $= -\frac{1}{2}\ln 2 + \frac{1}{2}\ln 4$ | A1 | from correct working o.e. e.g. $-\frac{1}{2}\ln(3+\cos(2\cdot\frac{\pi}{2})) + \frac{1}{2}\ln(3+\cos(2\cdot 0))$; o.e. required step for final A1, must have evaluated to 4 and 2 at this stage |
| $= \frac{1}{2}\ln\frac{4}{2}$ | | |
| $= \frac{1}{2}\ln 2$ * | A1 | **NB AG** |
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