`int_(pi/2)^pi sinx/(1+cos^2x) dx` Evaluate the definite integral

We have to evaluate the integral: `\int_{\pi/2}^{\pi}\frac{sinx}{1+cos^2x}dx`

Let `cosx=u`

So, `-sinx dx=du`

When `x=\pi/2, u=0`

         `x=\pi, u=-1`

So we have,

`\int_{\pi/2}^{\pi}\frac{sinx}{1+cos^2x}dx=\int_{0}^{-1}\frac{-du}{u^2+1}`

                       `=\int_{-1}^{0}\frac{du}{u^2+1}`

                        =arctan(0)-arctan(-1)

                        =-arctan(-1)

                        =arctan(1)

                        =`pi/4`