Answer
$$\frac{1}{2}\theta - \frac{1}{{20}}\sin 10\theta + C$$
Work Step by Step
$$\eqalign{
& \int {{{\sin }^2}5\theta } d\theta \cr
& {\text{trigonometric identity si}}{{\text{n}}^2}\phi = \frac{{1 - \cos 2\phi }}{2} \cr
& = \int {\frac{{1 - \cos 2\left( {5\theta } \right)}}{2}} d\theta \cr
& = \int {\frac{{1 - \cos 10\theta }}{2}} d\theta \cr
& = \int {\left( {\frac{1}{2} - \frac{{\cos 10\theta }}{2}} \right)} d\theta \cr
& = \int {\frac{1}{2}} d\theta - \int {\frac{{\cos 10\theta }}{2}} d\theta \cr
& {\text{find antiderivatives}} \cr
& = \frac{1}{2}\theta - \frac{1}{{2\left( {10} \right)}}\sin 10\theta + C \cr
& = \frac{1}{2}\theta - \frac{1}{{20}}\sin 10\theta + C \cr} $$
