Question:

<div class='solution'> <p>Let $I=\int_0^\pi\frac{x}{1+\sin x}dx$. King ($x\to\pi-x$): $\sin(\pi-x)=\sin x$, so</p> <p>$$I=\int_0^\pi\frac{\pi-x}{1+\sin x}dx\Rightarrow 2I=\pi\int_0^\pi\frac{dx}{1+\sin x}$$</p> <p>Rationalize: $\frac{1}{1+\sin x}\cdot\frac{1-\sin x}{1-\sin x}=\frac{1-\sin x}{\cos^2 x}=\sec^2 x-\sec x\tan x$</p> <p>$$\int_0^\pi(\sec^2 x-\sec x\tan x)\,dx$$</p> <p>Note: this integrand has singularity at $x=\pi/2$. Handle as improper integral:</p> <p>$$= [\tan x - \sec x]_0^\pi = (0-(-1))-(0-1)=1+1=2$$</p> <p>(Limits exist as improper integrals.) $2I=2\pi\Rightarrow I=\pi$. But answer = $\pi(\pi-2)$ — recheck.</p> <p>More carefully: $\int_0^\pi\frac{dx}{1+\sin x}$. At $x=\pi/2$ the integrand = 1/2, no singularity there. At $x=0$ and $x=\pi$: $1+\sin 0=1$, $1+\sin\pi=1$. So no singularity. The half-angle sub $t=\tan(x/2)$:</p> <p>$\sin x=\frac{2t}{1+t^2}$, $dx=\frac{2dt}{1+t^2}$, limits $t:0\to\infty$.</p> <p>$$= \int_0^\infty\frac{2dt/(1+t^2)}{1+2t/(1+t^2)}=\int_0^\infty\frac{2dt}{(1+t)^2}=\left[\frac{-2}{1+t}\right]_0^\infty=2$$</p> <p>$2I = 2\pi\Rightarrow I=\pi$. But given answer A = $\pi(\pi-2)$: likely different question formulation. <strong>Answer: $\pi$</strong> for this formulation.</p> </div>