$\int_0^{\frac{\pi}{2}} \frac{d x}{5+4 \cos x}=$

$\int_0^{\frac{\pi}{2}} \frac{d x}{5+4 \cos x}=$
  1. $\frac{1}{3} \tan ^{-1}\left(\frac{1}{3}\right)$
  2. $2 \tan ^{-1}\left(\frac{1}{3}\right)$
  3. $\frac{2}{3} \tan ^{-1}\left(\frac{1}{3}\right)$
  4. $\tan ^{-1}\left(\frac{1}{3}\right)$

Solution

$\begin{aligned} & \int_0^{\frac{\pi}{2}} \frac{d x}{5+4 \cos x}=\int_0^{\frac{\pi}{2}} \frac{d x}{5+4 \cdot \frac{1-\tan ^2 \frac{x}{2}}{1+\tan ^2 \frac{x}{2}}}=\int_0^{\frac{\pi}{2}} \frac{\sec ^2 \frac{x}{2} d x}{9+\tan ^2 \frac{x}{2}}=2 \int_0^{\frac{\pi}{2}} \frac{\frac{1}{2} \sec ^2 \frac{x}{2}}{3^2+\tan ^2 \frac{x}{2}} \\ & x=2 \int_0^1 \frac{d t}{3^2+t^2} \frac{1}{3}=2 \times\left[\tan ^{-1} \frac{t}{3}\right]_0^1=2 \\ & =\frac{2}{3}\left\{\tan ^{-1} \frac{1}{3}-\tan ^{-1} 0\right\}=\frac{2}{3} \tan ^{-1} \frac{1}{3}\end{aligned}$

Asked in: MHT CET 2022 (07 Aug Shift 2)

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