How To Evaluate $ \int \frac{\sin^2x\cdot\cos^2x}{(\sin^3x + \cos^3x)^2}\ dx $? $$
\int \frac{\sin^2x\cdot\cos^2x}{(\sin^3x + \cos^3x)^2}\ dx
$$
What I tried was 


*

*To convert $N^r$ into $sin2x$

*To Use Identity $(a+b)^3 = (a^3 + b^3)(a^2 + b^2 - ab) $
But none of them proved useful ?
How do I evaluate this Integral ?

Hint
Start with: $\sin x=\frac{\tan x}{\sec x}$ and $\sec^2 x =\tan^2 x +1$. Then,
$$\int \frac{\sin^2x\cdot\cos^2x}{(\sin^3x + \cos^3x)^2}\ dx = \int \sec^2x\,\frac{\tan^2x}{\left(\tan x+1\right)^2\left(\tan^2x-\tan x +1\right)^2}\,\mathrm{d}x.$$
Now use $u=\tan x$. Then use $t=\left(u+1\right)\left(u^2-u+1\right)$, sucht that
$$\int \frac{\sin^2x\cdot\cos^2x}{(\sin^3x + \cos^3x)^2}\ dx =\frac{1}{3}\int\frac{\mathrm{d}t}{t^2}.$$
