Evaluate:

$\sin ^{2} x=\frac{1-\cos 2 x}{2}$

$\therefore$ The given equation becomes,

$\Rightarrow \int \frac{1-\cos 2 \mathrm{~b}}{2} \mathrm{dx}$

We know $\int \cos a x d x=\frac{1}{a} \sin a x+c$

$\Rightarrow \frac{1}{2} \int \mathrm{dx}-\frac{1}{2} \int \cos (2 \mathrm{~b}) \mathrm{dx}$

$\Rightarrow \frac{x}{2}-\frac{1}{4 \mathrm{~b}} \sin (2 \mathrm{bx})+\mathrm{c}$