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$\displaystyle\int_0^\pi \dfrac{x}{1 + \sin x}\, dx = ?$
A$\pi$
B$\pi^2/2$
C$\pi/2$
D$2\pi$
Answer & Solution
Correct answer: A. $\pi$
King's rule $x \to \pi - x$: integrand becomes $(\pi - x)/(1 + \sin(\pi - x)) = (\pi - x)/(1 + \sin x)$. Adding to original: $2I = \pi\int_0^\pi dx/(1 + \sin x)$. Weierstrass $t = \tan(x/2)$: $dx/(1+\sin x) = 2dt/(1+t)^2$. $\int_0^\infty 2/(1+t)^2 dt = [-2/(1+t)]_0^\infty = 2$. So $2I = 2\pi$ ⇒ $I = \pi$.
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