How to evaluate $\int \frac { \sin x+\cos x }{ \sin^4 x+\cos^4x}\, dx$?

Question

How can one find $$\int \frac { \sin x+\cos x }{ \sin^4 x+\cos^4x}\, dx?$$

Answer 1

HINT:

$\sin^4x+\cos^4x=(\sin^2x+\cos^2x)^2-2\sin^2x\cos^2x=1-\dfrac{\sin^22x}2$

Set $\dfrac{d(\cos x+\sin x)}{dx}=-\sin x+\cos x=u\implies u^2=1-\sin2x$

Answer 2

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.

Answer 3

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}