Question

${\int }_{-\frac{\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}}\left[\frac{1}{\left(1+\cos x\right)}\right]dx=$

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

The correct option is C

$2$

Explanation for the correct option:

Finding the value for the given integral:

Let $I={\int }_{-\frac{\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}}\left[\frac{dx}{\left(1+\cos x\right)}\right]$

$$\begin{gathered} ={\int }_{-\frac{\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}}\left[\frac{1}{2{\cos }^2{\displaystyle \frac{x}{2}}}\right]dx\left[\mathrm{since}1+\mathrm{cosx}=2{\cos }^2\frac{x}{2}\right] \\ =\frac{1}{2}{\int }_{-\frac{\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}}se{c}^2\frac{x}{2}dx \\ =\frac{1}{2}\times \frac{1}{{\displaystyle \frac{1}{2}}}{\left[\tan \frac{x}{2}\right]}_{-\frac{\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}} \\ =\tan \frac{\mathrm{\pi }}{4}+\tan \frac{\mathrm{\pi }}{4} \\ =1+1 \\ =2 \end{gathered}$$

Hence, option C is the correct answer.