Evaluating ${\Large\int} _{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{\sin(x)+\cos(x)}{\sin^4(x)-4}dx$ This integral is giving me hard times, could anyone "prompt" a strategy about? I tried, resultless, parameterization and some change of variables.
$${\Large\int} _{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{\sin(x)+\cos(x)}{\sin^4(x)-4}dx$$

$$\int\dfrac{\sin x}{\sin^4 x-4}dx=-\int\dfrac{1}{(1-u^2)^2-4}du\\\int\dfrac{\cos x}{\sin^4 x-4}dx=\int\dfrac{du}{u^4-4}=0.25\int\dfrac{du}{u^2-2}-0.25\int\dfrac{du}{u^2+2}$$the first one is zero over a symmetric interval and for the second we have$$I=0.25\int_{-1}^{1}\dfrac{du}{u^2-2}-0.25\int_{-1}^{1}\dfrac{du}{u^2+2}=\dfrac{1}{4\sqrt 2}\ln\dfrac{\sqrt 2-1}{\sqrt 2+1}-{\sqrt{2}\over 4}\tan^{-1}\dfrac{1}{\sqrt 2}$$
