Evaluation of Trigonometric Integral 
Evaluation of $\displaystyle \int^{\frac{\pi}{4}}_{0}\frac{\sin^2 x\cdot \cos^2 x}{\sin^3 x+\cos^3 x}dx$

$\bf{My\; Try::} $ Let $$I= \int^{\frac{\pi}{4}}_{0}\frac{\sin^2 x\cdot \cos^2 x}{\sin^3 x+\cos^3 x}dx = \frac{1}{2}\int^{\frac{\pi}{4}}_{0}\frac{\sin^2 2x}{(\sin x+\cos x)(2-\sin 2x)}dx$$
So $$I = \frac{1}{2\sqrt{2}}\int^{\frac{\pi}{4}}_{0}\frac{\sin^2 2x}{\sin\left(x+\frac{\pi}{4}\right)(2-\sin 2x)}dx$$
Put $\displaystyle x+\frac{\pi}{4} = t,$ Then $dx = dt$ 
So $$I = \frac{1}{2\sqrt{2}}\int^{\frac{\pi}{2}}_{\frac{\pi}{4}}\frac{\cos^2 2t}{\sin t(2+\cos 2t)}dt=\frac{1}{2\sqrt{2}}\int^{\frac{\pi}{2}}_{\frac{\pi}{4}}\frac{\cos^2 2t-4+4}{\sin t(2+\cos 2t)}dt$$
So $$I = \frac{1}{2\sqrt{2}}\int^{\frac{\pi}{2}}_{\frac{\pi}{4}}\frac{\cos^2 2t-2}{\sin t}dt+\frac{2}{\sqrt{2}}\int^{\frac{\pi}{2}}_{\frac{\pi}{4}}\frac{1}{\sin t(2+\cos 2t)}dt$$
How can i solve it after that, Help required, Thanks

Hint
From where you left off:
The first integrand is the same as $$-\csc t-4\sin t+4\sin^3t$$
The second integral will be evaluated using the substitution $u=\cos t$ followed by a dose of partial fractions.
