Moderate -0.3 Part (i) is a straightforward application of the chain rule in reverse (power rule with linear substitution), requiring only routine technique. Part (ii) is a standard substitution u=1+2cos(x) leading to logarithmic integration, which is a textbook exercise for P3/C4 level. Both parts follow predictable patterns with no problem-solving insight required, making this slightly easier than average for an A-level question, though the 'show that' format and definite integral evaluation prevent it from being trivial.
dM1: Substitutes both \(0\) and \(\frac{\pi}{3}\) into expression of form \(b\ln(1+2\cos x)\) and subtracts; A1: \(\ln\frac{9}{4}\)
## Question 3:
### Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\int(2x-5)^7\,dx = \frac{(2x-5)^8}{16}+c$ | M1 | Achieves $a(2x-5)^8$ where $a$ is a constant, or equivalently $au^8$ with $u=(2x-5)$ |
| $\frac{(2x-5)^8}{16}+c$ | A1 | Exact simplified equivalent such as $\frac{1}{16}(2x-5)^8+c$; **+c must be present** |
### Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\int\frac{4\sin x}{1+2\cos x}\,dx = -2\ln(1+2\cos x)\ (+c)$ | M1 A1 | M1: Achieves $b\ln(1+2\cos x)$ or $b\ln|1+2\cos x|$ where $b$ is constant; A1: Achieves $-2\ln(1+2\cos x)$; no need for $+c$ |
| $\left[-2\ln(1+2\cos x)\right]_0^{\pi/3} = -2\ln 2+2\ln 3 = \ln\frac{9}{4}$ | dM1 A1 | dM1: Substitutes both $0$ and $\frac{\pi}{3}$ into expression of form $b\ln(1+2\cos x)$ and subtracts; A1: $\ln\frac{9}{4}$ |
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