Show that $\int_0^{\frac{\pi }{6}} {x\sin 2x{\text{d}}x = \frac{{\sqrt 3 }}{8} - \frac{\pi }{{24}}}$.

Using integration by parts     (M1)

$u = x{\text{, }}\frac{{{\text{d}}u}}{{{\text{d}}x}} = 1,{\text{ }}\frac{{{\text{d}}v}}{{{\text{d}}x}} = \sin 2x{\text{ and }}v = - \frac{1}{2}\cos 2x$     (A1)

$\left[ {x\left( { - \frac{1}{2}\cos 2x} \right)} \right]_0^{\frac{\pi }{6}} - \int_0^{\frac{\pi }{6}} {\left( { - \frac{1}{2}\cos 2x} \right){\text{d}}x}$     A1

$= \left[ {x\left( { - \frac{1}{2}\cos 2x} \right)} \right]_0^{\frac{\pi }{6}} + \left[ {\frac{1}{4}\sin 2x} \right]_0^{\frac{\pi }{6}}$     A1

Note: Award the A1A1 above if the limits are not included.

$\left[ {x\left( { - \frac{1}{2}\cos 2x} \right)} \right]_0^{\frac{\pi }{6}} = - \frac{\pi }{{24}}$     A1

$\left[ {\frac{1}{4}\sin 2x} \right]_0^{\frac{\pi }{6}} = \frac{{\sqrt 3 }}{8}$     A1

$\int_0^{\frac{\pi }{6}} {x\sin 2x{\text{d}}x = \frac{{\sqrt 3 }}{8} - \frac{\pi }{{24}}}$     AG     N0

Note: Allow FT on the last two A1 marks if the expressions are the negative of the correct ones.

[6 marks]

This question was reasonably well done, with few candidates making the inappropriate choice of u and $\frac{{{\text{d}}v}}{{{\text{d}}x}}$. The main source of a loss of marks was in finding v by integration. A few candidates used the double angle formula for sine, with poor results.