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

Question

$$ \int \frac{\sin^2x\cdot\cos^2x}{(\sin^3x + \cos^3x)^2}\ dx $$

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 ?

Answer 1

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

$$\int \frac{\sin^2x\cdot\cos^2x}{(\sin^3x + \cos^3x)^2}\ dx =\frac{1}{3}\int\frac{\mathrm{d}t}{t^2}.$$