Evaluating $\int_{0}^{\frac{\pi}{4}}\frac{\sin(4x)}{\cos^{4}x+\sin^{4}x}\,dx$

Question

Calculate

$$I=\int_{0}^{\frac{\pi}{4}}\frac{\sin(4x)}{\cos^{4}x+\sin^{4}x}\,dx$$

I tried to split the integral in a way to have integral like this:

$I=\int_{0}^{\frac{\pi}{2}}fdx-\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}fdx$ but I get stuck.How to start?

Answer 1

You have\begin{align}\frac{\sin(4x)}{\cos^4x+\sin^4x}&=\frac{2\sin(2x)\cos(2x)}{(\cos^2x+\sin^2x)^2-2\sin^2(x)\cos^2(x)}\\&=\frac{2\sin(2x)\cos(2x)}{1-\frac12\sin^2(2x)}\\&=-\frac{\left(1-\frac12\sin^2(2x)\right)'}{1-\frac12\sin^2(2x)}.\end{align}Can you take it from here?

Answer 2

Hint: Write your integrand in the form $$8\,{\frac {\sin \left( x \right) \left( \cos \left( x \right) \right) ^{3}}{2\, \left( \cos \left( x \right) \right) ^{4}-2\, \left( \cos \left( x \right) \right) ^{2}+1}}-4\,{\frac {\sin \left( x \right) \cos \left( x \right) }{2\, \left( \cos \left( x \right) \right) ^{4}-2\, \left( \cos \left( x \right) \right) ^{2}+ 1}} $$