Evaluating ${\Large\int} _{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{\sin(x)+\cos(x)}{\sin^4(x)-4}dx$

Question

This integral is giving me hard times, could anyone "prompt" a strategy about? I tried, resultless, parameterization and some change of variables.

$${\Large\int} _{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{\sin(x)+\cos(x)}{\sin^4(x)-4}dx$$

Answer 1

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

Answer 2

Following a suggestion from the comments, you can reduce your integral to$$\int_{-\frac\pi2}^{\frac\pi2}\frac{\cos x}{\sin^4(x)-4}\,\mathrm dx.$$Now, do the substitution $\sin x=t$ and $\cos x\,\mathrm dx=\mathrm dt$.

Answer 3

Remove the odd part and then set $t = \sin x$.

$$\int_{-\frac\pi2}^{\frac\pi2}\frac{\cos x}{\sin^4(x)-4}\,dx = \int_{-1}^1 \frac{dt}{t^4-4} = \frac14\int_{-1}^1 \frac{dt}{t^2-2} - \frac14\int_{-1}^1 \frac{dt}{t^2+2}$$

Can you finish?