## Question 4:

| Answer | Marks | Guidance |
|--------|-------|----------|
| $u=x$, $du/dx=1$, $dv/dx = \cos\frac{1}{2}x$, $v=2\sin\frac{1}{2}x$ | M1 | Correct $u$, $u'$, $v$, $v'$. But allow $v$ to be any multiple of $\sin\frac{1}{2}x$. M0 if $u=\cos\frac{1}{2}x$, $v'=x$ |
| $\int_0^{\pi/2} x\cos\frac{1}{2}x\,dx = \left[2x\sin\frac{1}{2}x\right]_0^{\pi/2} - \int_0^{\pi/2} 2\sin\frac{1}{2}x\,dx$ | A1ft | Consistent with their $u$, $v$ |
| $= \left[2x\sin\frac{1}{2}x + 4\cos\frac{1}{2}x\right]_0^{\pi/2}$ | A1 | $2x\sin\frac{1}{2}x + 4\cos\frac{1}{2}x$ oe (no ft) |
| $= \pi\sin\frac{\pi}{4} + 4\cos\frac{\pi}{4} - (2\cdot0\cdot\sin 0 + 4\cos 0)$ | M1 | Substituting correct limits into correct expression. Can be implied by one correct intermediate step |
| $= \pi\cdot\frac{1}{\sqrt{2}} + 4\cdot\frac{1}{\sqrt{2}} - 4$ | | |
| $= \frac{\sqrt{2}}{2}\pi + 2\sqrt{2} - 4$ | A1cao | **NB AG** |
| **[5]** | | |

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