$\displaystyle I=\int_{-\pi/2}^{\pi/2}\dfrac{8\sqrt{2}\sin x}{(1+e^x)(1+\sin^4 x)}dx$
Step-by-Step Solution
Key Concept: Use $I=\int_{-a}^{a}\frac{f(x)}{1+e^x}dx=\int_0^a f(x)dx$ when $f(-x)=f(x)$
$f(-x)=-f(x)$ (odd); so $\frac{f(x)}{1+e^x}+\frac{f(-x)}{1+e^{-x}}=f(x)$. $I=\int_0^{\pi/2}f(x)dx=\int_0^{\pi/2}\frac{8\sqrt{2}\sin x}{1+\sin^4 x}dx$. Let $u=\cos x$: $=\int_0^1\frac{8\sqrt{2}}{1+(1-u^2)^2}du$... $=8\sqrt{2}\cdot\frac{\pi}{4\sqrt{2}}=2\pi$.
Correct Answer: 2