Integration challenge 2

Note that :

$\displaystyle \frac{1}{1+\cos 2x } - \frac{1}{1+\sin 2x } = \frac{\sin 2x - \cos 2x }{(1+\cos 2x)(1+\sin 2x) }$

$\displaystyle = \frac{\sin 2x - \cos 2x }{(2\cos^2x)(1+\sin 2x) }$

Hence:

$\displaystyle \int_0^{\frac{\pi}{4}} x^2 \frac{\sin 2x - \cos 2x }{(\cos^2x)(1+\sin 2x) }dx$

$\displaystyle = \int_0^{\frac{\pi}{4}} 2x^2 ( \frac{1}{1+\cos 2x } - \frac{1}{1+\sin 2x } ) dx$

$\displaystyle = \int_0^{\frac{\pi}{4}} \frac{2x^2}{1+\cos 2x } dx - \int_0^{\frac{\pi}{4}} \frac{2x^2}{1+\sin 2x } dx$

Use the substitution $\; x \to (\frac{\pi}{4} - x) \;$ in the second integral :

$\displaystyle = \int_0^{\frac{\pi}{4}} \frac{2x^2}{1+\cos 2x } dx - \int_0^{\frac{\pi}{4}} \frac{2(\frac{\pi}{4} - x)^2}{1+\cos 2x } dx$

$\displaystyle = 2 \int_0^{\frac{\pi}{4}} \frac{x^2-(\frac{\pi}{4} - x)^2 }{1+\cos 2x } dx$

$\displaystyle = \frac{\pi}{2} \int_0^{\frac{\pi}{4}} \frac{2x-\frac{\pi}{4}}{1+\cos 2x } dx$

$\displaystyle = \frac{\pi}{2} \int_0^{\frac{\pi}{4}} (x-\frac{\pi}{8}) \sec^2x dx$

The rest is easily manipulated by integration by parts... To get the value:

$\displaystyle \boxed{ = \frac{\pi ^2}{16} - \frac{\pi^1}{4} \ln (2) }$

So the answer is : $2+16+1+4+2 = 25$ .