prove it.....if u can

we get

2I = 2pixcosx/(1+sinx)^2 - pi^2cosx/(1+sinx)^2 .........[put y = pi-x and add with org. eq]

again

let

I1 = 2pixcosx/(1+sinx)^2 I2=pi^2cosx/(1+sinx)^2

2I1 = 2pi^2cosx/(1+sinx)^2 ............

now we can solve I1 and I2 by puttin sinx = t ....

havnt really solved .... try it out ....

Let
I=\int_0^\pi \dfrac{x^2\cos x\ \mathrm dx}{(1+\sin x)^2}
Then
I=\int_0^\pi \dfrac{-(\pi-x)^2\cos x\ \mathrm dx}{(1+\sin x)^2}
Hence,
2I=\int_0^\pi \dfrac{(x^2-(\pi-x)^2)\cos x\ \mathrm dx}{(1+\sin x)^2}=\int_0^\pi \dfrac{\pi(2x-\pi)\cos x\ \mathrm dx}{(1+\sin x)^2}=2\pi\int_0^\pi \dfrac{x\cos x\ \mathrm dx}{(1+\sin x)^2}-\pi^2\int_0^\pi \dfrac{\cos x\ \mathrm dx}{(1+\sin x)^2}=2\pi I_1 -\pi^2 I_2
The second integeral is obviously 0. So we get
2I = 2\pi I_1\quad\Rightarrow\ I = \pi I_1
Notice that
\dfrac{\cos x\ \mathrm dx}{(1+\sin x)^2} = \dfrac{\mathrm d(1+\sin x)}{(1+\sin x)^2}=-\mathrm d\left(\dfrac{1}{1+\sin x}\right)
So that
\int\dfrac{\cos x\ \mathrm dx}{(1+\sin x)^2} =- \dfrac{1}{1+\sin x}
Hence,
I_1=\left[-\dfrac{x}{1+\sin x}\right]_0^\pi+\int_0^\pi \dfrac{\mathrm dx}{1+\sin x}=-\pi + 2
Thus
I=\pi(2-\pi)