Q115 / 160 · AP EAMCET 2024 (23 May Shift 1) · Mathematics
If $\int_0^\pi \frac{x \sin x}{4 \cos ^2 x+3 \sin ^2 x} d x=$
- Correct answerA$\frac{\pi^2}{6 \sqrt{3}}$
- B$\frac{\pi}{3 \sqrt{3}}$
- C$\frac{\pi^2}{3 \sqrt{3}}$
- D$\sqrt{3} \pi^2$
Solution
$I=\int_0^\pi \frac{x \sin x}{4 \cos ^2 x+3 \sin ^2 x} d x$
$\Rightarrow I=\int_0^\pi \frac{x \sin x}{\cos ^2 x+3} d x$ ....(i)
$\Rightarrow I=\int_0^\pi \frac{(\pi-x) \sin x}{\cos ^2 x+3} d x$ ....(ii)
Adding (i) and (ii)
$2 I=\pi \int_0^\pi \frac{\sin x}{\cos ^2 x+3} d x$
Let $\cos x=t \Rightarrow-\sin x d x=d t$ $\begin{aligned} & \therefore 2 I=-\pi \int_1^{-1} \frac{1}{t^2+3} d t \Rightarrow 2 I=-\pi \int_{-1}^1 \frac{1}{t^2+3} d t \\ & \Rightarrow 2 I=\frac{\pi}{\sqrt{3}}\left[\tan ^{-1} \frac{t}{\sqrt{3}}\right]_{-1}^1 \\ & \Rightarrow 2 I=\frac{\pi}{\sqrt{3}}\left[\frac{\pi}{6}+\frac{\pi}{6}\right] \Rightarrow I=\frac{\pi^2}{6 \sqrt{3}} \end{aligned}$
Let $\cos x=t \Rightarrow-\sin x d x=d t$ $\begin{aligned} & \therefore 2 I=-\pi \int_1^{-1} \frac{1}{t^2+3} d t \Rightarrow 2 I=-\pi \int_{-1}^1 \frac{1}{t^2+3} d t \\ & \Rightarrow 2 I=\frac{\pi}{\sqrt{3}}\left[\tan ^{-1} \frac{t}{\sqrt{3}}\right]_{-1}^1 \\ & \Rightarrow 2 I=\frac{\pi}{\sqrt{3}}\left[\frac{\pi}{6}+\frac{\pi}{6}\right] \Rightarrow I=\frac{\pi^2}{6 \sqrt{3}} \end{aligned}$
Hence, the correct answer is option: 'A'.