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>