<p>Evaluate \(\displaystyle\int_0^{\pi}\frac{\theta\sin\theta}{1+\cos^2\theta}\,d\theta\) [JEE Advanced 1999]</p>
Step-by-Step Solution
Key Concept: King's rule: I = \int_0^\pi (\pi-\theta)sin\theta/(1+cos^2\theta)d\theta. Add: 2I = \pi\int_0^\pi sin\theta/(1+cos^2\theta)d\theta = \pi \cdot [arctan(cos\theta)... wait: -arctan(cos\theta)]_0^\pi = \pi \cdot (-(-\pi/4)+(\pi/4))... Let t=cos\theta: \pi\int₋_1^1dt/(1+t^2)=\pi \cdot \pi/2. So I=\pi^2/4.
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<p>Let \(I=\int_0^\pi\frac{\theta\sin\theta}{1+\cos^2\theta}d\theta\). King (\(\theta\to\pi-\theta\)): \(\sin(\pi-\theta)=\sin\theta\), \(\cos(\pi-\theta)=-\cos\theta\), \(\cos^2(\pi-\theta)=\cos^2\theta\).</p>
<p>\[I=\int_0^\pi\frac{(\pi-\theta)\sin\theta}{1+\cos^2\theta}d\theta\]</p>
<p>Add: \(2I=\pi\int_0^\pi\frac{\sin\theta}{1+\cos^2\theta}d\theta\).</p>
<p>Let \(t=\cos\theta\), \(dt=-\sin\theta\,d\theta\):</p>
<p>\[\int_0^\pi\frac{\sin\theta}{1+\cos^2\theta}d\theta=\int_1^{-1}\frac{-dt}{1+t^2}=\int_{-1}^1\frac{dt}{1+t^2}=\frac{\pi}{2}\]</p>
<p>\[2I=\frac{\pi^2}{2}\Rightarrow I=\boxed{\frac{\pi^2}{4}}\]</p>
Correct Answer: A