# integral calculate by complex analysis methods

Calculate using methods from comples analysis. $$\int_0^{2\pi} \,\sin ^{2n} \phi\, d\phi$$

So this is how I started:

$$\sin^{2n} \phi = \left[\frac{e^{i \phi}-e^{-i \phi}}{2i}\right]^{2n} = \left[\frac{e^{2i \phi}+e^{-2i \phi}-2}{-4}\right]^{n}$$

But I don't know what to do next?

If you are using complex techniques then use the substitution $z=e^{i\phi}$ and the your integral in this case becomes the contour integral

$$\frac{1}{(2i)^{2n}}\int_{|z|=1} \left(z+\frac{1}{z}\right)^{2n}\frac{dz}{iz} = \frac{1}{(-4)^{n}} \int_{|z|=1} \frac{(z^2+1)^{2n}}{z^{2n}}\frac{dz}{iz}.$$

Now you need to know your poles and then find the residue.

Note:

1) Use the binomial theorem to find the residue.

$$(a+b)^{r} = \sum_{m=0}^{r} { r\choose m } a^{m}b^{r-m}.$$