Evaluate the following integral: `int_0^pi |sin^2x|dx`

Explanation

`I=int_{0}^{pi} sin ^{2} x d x`

`
int_{0}^{pi} sin ^{2} x d x=int_{0}^{pi}(frac{1-cos 2 x}{2}) d x(because sin ^{2} x=frac{1-cos 2 x}{2})`

`
=frac{1}{2}[int_{0}^{pi} d x-int_{0}^{pi}(cos 2 x) d x] `

`
=frac{1}{2}[x-frac{sin 2 x}{2}]_{0}^{pi}`

`
=frac{1}{2}[(pi-0)-frac{(sin 2 pi-sin 0)}{2}]=frac{pi}{2}`

`