Q65 / 125 · BITSAT 2021 · Mathematics
Evaluate: $\int_{0}^{\pi} \frac{1}{5+4 \cos x} d x$
- Correct answerA$\frac{\pi}{3}$
- B$\frac{\pi}{2}$
- C$\frac{\pi}{4}$
- D$\frac{\pi}{6}$
Solution
We have, $\int_{0}^{\pi} \frac{1}{5+4 \cos x} d x$
$=\int_{0}^{\pi} \frac{1}{5+4\left(\frac{1-\tan ^{2} \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}\right)} d x$
$=\int_{0}^{\pi} \frac{1+\tan ^{2} \frac{x}{2}}{5\left(1+\tan ^{2} \frac{x}{2}\right)+4\left(1-\tan ^{2} \frac{x}{2}\right)} d x$
$=\int_{0}^{\pi} \frac{1+\tan ^{2} \frac{x}{2}}{9+\tan ^{2} \frac{x}{2}} \mathrm{dx}=\int_{0}^{\pi} \frac{\sec ^{2} \frac{x}{2}}{9+\tan ^{2} \frac{x}{2}} \mathrm{dx}$
Let $\tan \frac{x}{2}=t \Rightarrow \frac{1}{2} \sec ^{2} \frac{x}{2} d x=d t$
Also, $x=0 \Rightarrow t=0$ and $x=\pi \Rightarrow t=\infty$
$\therefore \mathrm{I}=\int_{0}^{\infty} \frac{\mathrm{dt}}{9+t^{2}}$
$\therefore \mathrm{I}=2 \int_{0}^{\infty} \frac{\mathrm{dt}}{9+t^{2}}$
$\therefore \mathrm{I}=\frac{2}{3}\left[\tan ^{-1} \frac{t}{3}\right]_{0}^{\infty}=\frac{2}{3}\left[\tan ^{-1} \infty-\tan ^{-1} 0\right]$
$=\frac{2}{3}\left(\frac{\pi}{2}-0\right)=\frac{\pi}{3}$
Hence, the correct answer is option: 'A'.