$\int_{0}^{\frac{\pi}{2}} \frac{d x}{1+\cos x}=$

$\int_{0}^{\frac{\pi}{2}} \frac{d x}{1+\cos x}=$
  1. -2
  2. 2
  3. 1
  4. -1

Solution

$\begin{aligned}{\text{Let}}1 &=\int_{0}^{\frac{\pi}{2}} \frac{\mathrm{dx}}{1+\cos \mathrm{x}}=\int_{0}^{\frac{\pi}{2}} \frac{\mathrm{dx}}{2 \cos ^{2} \frac{\mathrm{x}}{2}}=\frac{1}{2} \int_{0}^{\frac{\pi}{2}} \sec ^{2} \frac{\mathrm{x}}{2} \mathrm{dx} \\ &=\frac{1}{2}\left[\frac{\tan \frac{\mathrm{x}}{2}}{\left(\frac{1}{2}\right)}\right]_{0}^{\frac{\pi}{2}}=\tan \frac{\pi}{4}-\tan 0=1-0=1 \end{aligned}$

Asked in: MHT CET 2020 (14 Oct Shift 2)

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