The value of \(\int_{0}^{\pi}\displaystyle\frac{x\sin x}{3+\cos^{2}x}\,dx\) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\displaystyle\frac{\pi^{2}}{6\sqrt{3}}\)
Let
\( \displaystyle I = \int_{0}^{\pi}\frac{x\sin x}{3+\cos^{2}x}\,dx \qquad (1) \)
By the property \(\int_{0}^{a} f(x)\,dx = \int_{0}^{a} f(a-x)\,dx\), replace x by \(\pi - x\). Since \(\sin(\pi - x) = \sin x\) and \(\cos^{2}(\pi - x) = \cos^{2}x\),
\( \displaystyle I = \int_{0}^{\pi}\frac{(\pi - x)\sin x}{3+\cos^{2}x}\,dx \qquad (2) \)
Adding (1) and (2), the x terms cancel:
\( \displaystyle 2I = \pi\int_{0}^{\pi}\frac{\sin x}{3+\cos^{2}x}\,dx \)
Let \(t = \cos x\), so \(dt = -\sin x\,dx\); the limits 0 and \(\pi\) become 1 and \(-1\):
\( \displaystyle 2I = \pi\int_{-1}^{1}\frac{dt}{3+t^{2}} \)
\( \displaystyle = \pi\cdot\frac{1}{\sqrt{3}}\left[\tan^{-1}\frac{t}{\sqrt{3}}\right]_{-1}^{1} \)
\( \displaystyle = \frac{\pi}{\sqrt{3}}\left(\frac{\pi}{6} + \frac{\pi}{6}\right) = \frac{\pi^{2}}{3\sqrt{3}} \)
\( \displaystyle I = \frac{\pi^{2}}{6\sqrt{3}} \)
Why the other options are incorrect
Option (1) \(\displaystyle\frac{\pi^{2}}{3\sqrt{3}}\) — Incorrect. This is 2I; it forgets to divide by 2 at the end.
Option (2) \(\displaystyle\frac{\pi}{6\sqrt{3}}\) — Incorrect. It loses the factor \(\pi\) that comes out of \(x + (\pi - x) = \pi\).
Option (4) \(\displaystyle\frac{\pi^{2}}{12\sqrt{3}}\) — Incorrect. It takes the t-integral over \([0, 1]\) only, which is half of the integral over \([-1, 1]\).