allery$\displaystyle\int_0^\pi \dfrac{x\sin x}{1 + \cos^2 x}\, dx = ?$
A$\pi^2/4$
B$\pi/4$
C$\pi^2/8$
D$\pi$
Answer & Solution
Correct answer: A. $\pi^2/4$
Use $\int_0^\pi x f(\sin x)\, dx = (\pi/2)\int_0^\pi f(\sin x)\, dx$.
Here $f(\sin x) = \sin x/(1 + \cos^2 x)$.
Now sub $u = \cos x$, $du = -\sin x\, dx$: $\int_0^\pi \sin x/(1+\cos^2 x)\, dx = \int_{-1}^{1} du/(1+u^2) = [\arctan u]_{-1}^1 = \pi/2$.
**Final:** $(\pi/2)(\pi/2) = \pi^2/4$.
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