Answer
$$\frac{1}{2}\pi $$
Work Step by Step
$$\eqalign{
& \int_0^\pi {{{\sin }^2}5r} dr \cr
& {\text{use si}}{{\text{n}}^2}\theta = \frac{1}{2} - \frac{1}{2}\cos 2\theta \cr
& \int_0^\pi {{{\sin }^2}5r} dr = \int_0^\pi {\left( {\frac{1}{2} - \frac{1}{2}\cos 2\left( {5r} \right)} \right)} dr \cr
& = \int_0^\pi {\left( {\frac{1}{2} - \frac{1}{2}\cos 10r} \right)} dr \cr
& {\text{integrating}} \cr
& = \left( {\frac{1}{2}r - \frac{1}{{2\left( {10} \right)}}\sin 10r} \right)_0^\pi \cr
& = \frac{1}{2}\left( {r - \frac{1}{{10}}\sin 10r} \right)_0^\pi \cr
& {\text{Evaluating, we get:}} \cr
& = \frac{1}{2}\left( {\pi - \frac{1}{{10}}\sin 10\pi } \right) - \frac{1}{2}\left( {0 - \frac{1}{{10}}\sin 10\left( 0 \right)} \right) \cr
& = \frac{1}{2}\pi \cr} $$
