Hi,
I've attached a problem I've come across and was wondering if anyone could give me a hand?
Cheers.
You need to evaluate $\displaystyle I_k = \int_0^{\pi/2}\sin^{2k}\phi\,d\phi.$ Integrate by parts:
$\displaystyle \begin{aligned}I_k &= \int_0^{\pi/2}\sin\phi\sin^{2k-1}\phi\,d\phi \\ &= \Bigl[-\cos\phi\sin^{2k-1}\phi\Bigr]_0^{\pi/2} + \int_0^{\pi/2}(2k-1)\cos^2\phi\sin^{2k-2}\phi\,d\phi \\ &= \int_0^{\pi/2}(2k-1)(1-\sin^2\phi)\sin^{2k-2}\phi\,d\phi = (2k-1)(I_{k-1}-I_k).\end{aligned}$
Therefore $\displaystyle I_k = \frac{2k-1}{2k}I_{k-1} = \frac{k-\frac12}kI_{k-1}.$
Now check that $\displaystyle I_1=\pi/4$ and use induction to show that $\displaystyle I_k = \frac{\pi}2\,\frac{\bigl(\frac12\bigr)_k}{k!}.$