definite integral $\int_{0}^{\frac{\pi}{4}} \frac{\sin^2x\cos^2x}{\sin^3x+\cos^3x}dx$ 
$$\int_{0}^{\frac{\pi}{4}} \frac{\sin^2x\cos^2x}{\sin^3x+\cos^3x}dx$$

I tried $2$ or $3$ trigonometric transformations but it did not work. One of them is as follows 
$$\frac{\sin^2x\cos^2x}{(\sin x+\cos x)(1-\sin x\cos x)}$$ after that I am not able to figure out what to do. If I use double angle formula then 
$$\frac{\frac{\sin2x}{4}}{(\sin x+\cos x)\left(1-\frac{\sin2x}{2}\right)}$$ again i am clueless

There's a small mistake in your attempt. It's $\frac{\frac{\sin^{2}2x}{4}}{(\sin x+\cos x)\left(1-\frac{\sin2x}{2}\right)}$
Note that we can write this as 
$\frac{\frac{\sin^{2}2x}{4}}{\sqrt2\sin(\frac{\pi}{4}+x)\left(1-\frac{\sin2x}{2}\right)}$.
Hint: Replacing $x$ with $\frac{\pi}{4}-x$ seems like a good idea here. 
