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
 A: Using identities
$$\sin x+\cos x=\sqrt{2}\cos(x-\dfrac{\pi}{4})$$
$$\sin x\cos x=\cos^2(x-\dfrac{\pi}{4})-\dfrac12$$
and then substitution $\dfrac{\pi}{4}-x=u$ gives
\begin{align}
I
&= \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 x+\cos x)(1-\sin x\cos x)}dx \\
&= \dfrac{1}{2\sqrt{2}}\int_{0}^{\frac{\pi}{4}}\frac{(2\cos^2u-1)^2}{\cos u(3-2\cos^2u)}\ du \\
&= \dfrac{1}{2\sqrt{2}}\int_{0}^{\frac{\pi}{4}}\frac{(2\cos^2u-1)^2}{\cos^2u(3-2\cos^2u)}\cos u\  du
\end{align}
now let $\sin u=w$ therefore
\begin{align}
I
&= \dfrac{1}{2\sqrt{2}}\int_{0}^{\frac{\sqrt{2}}{2}} \frac{(2w^2-1)^2}{(1-w^2)(2w^2+1)}\ dw \\
&= \dfrac{1}{2\sqrt{2}}\int_{0}^{\frac{\sqrt{2}}{2}} -2+\frac13\frac{1}{1-w^2}+\frac83\frac{1}{2w^2+1}\ dw \\
&= \color{blue}{-\dfrac12+\dfrac{\pi}{6}+\dfrac{\sqrt{2}}{12}\operatorname{arccoth}\sqrt{2}}
\end{align}
A: 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. 
A: Nosrati's idea is good. However his second identity is wrong. Using identities
$$\sin x+\cos x=\sqrt{2}\cos(\dfrac{\pi}{4}-x), \sin x\cos x=\frac12\big[2\cos^2(\dfrac{\pi}{4}-x)-1\big]$$
and substitution $x\to\dfrac{\pi}{4}-x$ gives
\begin{eqnarray}
I&=&\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 x+\cos x)(1-\sin x\cos x)}dx\\
&=&\frac14\int_{0}^{\frac{\pi}{4}}\frac{\sin^2(2x)}{\sqrt2\cos(\frac{\pi}{4}-x)(1-\frac12[\cos^2(x-\dfrac{\pi}{4})-1])}dx\\
&=&\int_{0}^{\frac{\pi}{4}}\frac{\frac14\big[2\cos^2(\dfrac{\pi}{4}-x)-1\big]^2}{\sqrt{2}\cos(\dfrac{\pi}{4}-x)(1-\frac12\big[2\cos^2(\dfrac{\pi}{4}-x)-1\big])}dx\\
&=&\frac1{2\sqrt2}\int_{0}^{\frac{\pi}{4}}\frac{(1-2\cos^2x)^2}{\cos x(3-2\cos^2x)}dx\\
&=&\frac1{6\sqrt2}\int_{0}^{\frac{\pi}{4}}\bigg(\frac{(1-2\cos^2x)^2}{\cos x}+\frac{2\cos x(1-2\cos^2x)^2}{3-2\cos^2x}\bigg)dx.
\end{eqnarray}
Noting that
\begin{eqnarray}
\int\frac{(1-2\cos^2x)^2}{\cos x}dx&=&\int(4 \cos ^3(x)-4 \cos (x)+\sec (x))dx\\
&=&\frac13\sin(3x)-\sin x+\ln(\sec x+\tan x)+C\\
\int\frac{2\cos x(1-2\cos^2x)^2}{(3-2\cos^2x)}dx&=&2\int\frac{(2\sin^2x-1)^2}{1+2\sin^2x}d\sin x\\
&=&2\int\bigg(-3+2\sin^2x+\frac{4}{1+2\sin^2x}\bigg)d\sin x\\
&=&2\bigg(-3\sin x+\frac{2}{3}\sin^3x+2\sqrt2\arctan(\sqrt2\sin x)+C\bigg).
\end{eqnarray}
So
\begin{eqnarray}
I&=&\frac1{6\sqrt2}\bigg(\frac13\sin(3x)-\sin x+\ln(\sec x+\tan x)\\
&&+2\bigg(-3\sin x+\frac{2}{3}\sin^3x+2\sqrt2\arctan(\sqrt2\sin x)\bigg)\bigg)\bigg|_0^{\pi/4}\\
&=&\frac{1}{6\sqrt2}(-3\sqrt2+\pi\sqrt2+\ln(\sqrt2+1)).
\end{eqnarray}
A: Hint: Show that your integrand is equal to $$\frac{\cos (x) \cot (x)}{\cot ^3(x)+1}$$
