How to evaluate $\int \frac { \sin x+\cos x }{ \sin^4 x+\cos^4x}\, dx$? How can one find $$\int  \frac { \sin x+\cos x }{ \sin^4 x+\cos^4x}\, dx?$$
 A: HINT:
$\sin^4x+\cos^4x=(\sin^2x+\cos^2x)^2-2\sin^2x\cos^2x=1-\dfrac{\sin^22x}2$
Set $\dfrac{d(\cos x+\sin x)}{dx}=-\sin x+\cos x=u\implies u^2=1-\sin2x$
A: Split the integral as
$$
\int\frac {\cos x}{\sin^4 x + (1 - \sin^2 x)^2} dx 
+ \int \frac{\sin x}{(1 - \cos^2 x)^2 + \cos^4 x} dx
$$
Compute the integrals with the substitutions $u = \sin x$ and $u = \cos x$ respectively.
A: Let
\begin{align}
 I &= \int {\frac{{\sin x + \cos x}}{{\sin ^4 x + \cos ^4 x}}dx}  \\ 
  &= \int {\frac{{\sin x}}{{\sin ^4 x + \cos ^4 x}}dx}  + \int {\frac{{\cos x}}{{\sin ^4 x + \cos ^4 x}}dx}  \\ 
  &= \int {\frac{{\sin x}}{{\sin ^4 x + \cos ^4 x}}dx}  + \int {\frac{{\sin \left( {{\textstyle{\pi  \over 2}} - x} \right)}}{{\sin ^4 x + \sin ^4 \left( {{\textstyle{\pi  \over 2}} - x} \right)}}dx} ,\,\,\,\,\,\, (\text{since}\,\, 
{\sin \left( {{\textstyle{\pi  \over 2}} - x} \right)}=\cos x)
\end{align}
Substituting $u=\frac{\pi}{2}-x$ in the second integral we get
\begin{align}
I&= \int {\frac{{\sin x}}{{\sin ^4 x + \cos ^4 x}}dx}  - \int {\frac{{\sin u}}{{\sin ^4 \left( {{\textstyle{\pi  \over 2}} + u} \right) + \sin ^4 u}}du} 
\\
 &= \int {\frac{{\sin x}}{{\sin ^4 x + \cos ^4 x}}dx}  - \int {\frac{{\sin u}}{{\cos ^4 u + \sin ^4 u}}du} ,\,\,\,\,\,\,\,\,\,\,\,\, (\text{since}\,\, 
{\sin \left( {{\textstyle{\pi  \over 2}} + u} \right)}=\cos u)
\\
&=0
\end{align}
