Answer
$$\text{a)} \displaystyle\int\sin^2x\ dx=\frac{1}{2}x-\frac{1}{4}\sin {2x}+C$$
$$\text{b)}\displaystyle{\int\sin^4x\ dx=-\frac{1}{8}\sin {2x}\sin^{2}x+\frac{3}{8}x-\frac{3}{16}\sin{2x}+C}$$
Work Step by Step
$\displaystyle{\int\sin^nx\ dx=-\frac{1}{n}\cos x\sin^{n-1}x+\frac{n-1}{n}\int \sin^{n-2}x\ dx}$
a)
n=2
$\displaystyle\int\sin^2x\ dx=-\frac{1}{2}\cos x\sin x+\frac{1}{2}\int \sin^{0}x\ dx\\
\displaystyle\int\sin^2x\ dx=-\frac{1}{2}\times\frac{1}{2}\times2\cos x\sin x+\frac{1}{2}\int 1\ dx\\
\displaystyle\int\sin^2x\ dx=-\frac{1}{4}\sin {2x}+\frac{1}{2}x+C\\
\displaystyle\int\sin^2x\ dx=\frac{1}{2}x-\frac{1}{4}\sin {2x}+C$
b)
$I=\displaystyle\int\sin^4x\ dx\\
I=\displaystyle -\frac{1}{4}\cos x\sin^{3}x+\frac{3}{4}\int \sin^{2}x\ dx$
(From a)
$I=-\frac{1}{4}\cos x\sin^{3}x+\frac{3}{4}\left(\frac{1}{2}x-\frac{1}{4}\sin {2x}\right)+C\\
I=-\frac{1}{4}\times\frac{1}{2}\times2\cos x\sin x\sin^{2}x+\frac{3}{8}x-\frac{3}{16}\sin {2x}+C\\
I=-\frac{1}{8}\sin {2x}\sin^{2}x+\frac{3}{8}x-\frac{3}{16}\sin {2x}+C$
