Evaluate $\int_0^{\frac{\pi}{2}}\frac{\sin x\cos x}{\sin^4x+\cos^4x}dx$ 
Evaluate
  $$
\int_0^{\frac{\pi}{2}}\frac{\sin(x)\cos(x)}{\sin^4(x)+\cos^4(x)}dx
$$

I used the substitution $\sin x =t$, then I got the integral as $$\int_0^1 \frac{t}{2t^4-2t^2+1}dt $$ 
After that I don't know how to proceed. Please help me with this.
 A: Do  some trigonometry first:
\begin{align}
\frac{\sin x\cos x}{\sin^4x+\cos^4x}&=\frac{\tfrac12\sin 2x}{(\sin^2x+\cos^2x)^2-2\sin^2x\cos^2x}=\frac{\tfrac12\sin 2x}{1-\frac12\sin^22x}\\&=\frac{\sin 2x}{2-\sin^22x}=\frac{\sin 2x}{1+\cos^22x}.
\end{align}
Next use substitution: set $\;u=\cos 2x$, $\;\mathrm d u=-2\sin 2x\,\mathrm d x$. 
A: Hint:
$$\dfrac{\sin x\cos x}{\sin^4x+\cos^4x}=\dfrac{\tan x\sec^2x}{\tan^4x+1}$$
Set $\tan^2x=y$
OR $$\dfrac{\sin x\cos x}{\sin^4x+\cos^4x}=\dfrac{\cot x\csc^2x}{\cot^4x+1}$$
Set $\cot^2x=u$
A: If you want to continue your solution, then with substitution $t^2=u$
$$I=\int_0^1\dfrac{t}{2t^4-2t^2+1}dt=\dfrac12\int_0^1\dfrac{1}{2u^2-2u+1}du=\int_0^1\dfrac{1}{(2u-1)^2+1}du$$
and then with substitution $2u-1=w$
$$I=\dfrac12\int_{-1}^1\dfrac{1}{w^2+1}dw=\color{blue}{\dfrac{\pi}{4}}$$
A: Letting $u=\tan x$, one has
\begin{eqnarray}
&&\int_0^{\frac{\pi}{2}}\frac{\sin(x)\cos(x)}{\sin^4(x)+\cos^4(x)}dx\\
&=&\int_0^{\frac{\pi}{2}}\frac{\tan(x)\sec^2(x)}{\tan^4(x)+1}dx\\
&=&\int_0^\infty\frac{u}{u^4+1}du\\
&=&\frac12\int_0^\infty\frac{1}{u^2+1}du\\
&=&\frac12\arctan（u)\bigg|_0^\infty\\
&=&\frac\pi4.
\end{eqnarray}
A: Perform the change of variable $y=\sin^2 x$,
$\begin{align}J&=\int_0^{\frac{\pi}{2}}\frac{\sin(x)\cos(x)}{\sin^4(x)+\cos^4(x)}dx\\
&=\frac{1}{2}\int_0^1 \frac{1}{x^2+(1-x)^2}\,dx\\
&=\frac{1}{2}\int_0^1 \frac{1}{2x^2-2x+1}\,dx\\
&=\int_0^1 \frac{1}{(2x-1)^2+1}\,dx\\
\end{align}$
Perform the change of variable $y=2x-1$,
$\begin{align}
J&=\frac{1}{2}\int_{-1}^1 \frac{1}{x^2+1}\,dx\\
&=\frac{1}{2}\Big[\arctan x\Big]_{-1}^1\\
&=\frac{1}{2}\times\frac{\pi}{2}\\
&=\boxed{\frac{\pi}{4}}
\end{align}$
