Question

${\int }_{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\cos e{c}^{2}xdx=$

• A. $-1$
• B. 1
• C. $0$
• D. $\frac{1}{2}$

Answer 1

The correct option is B

1

Explanation for the correct option:

Finding the value of the given integral:

Let $I={\int }_{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\cos e{c}^{2}xdx$

$$\begin{gathered} ={\left[-cotx\right]}_{\frac{\mathrm{\pi }}{4}}^{\frac{\mathrm{\pi }}{2}}\left[\mathrm{since}\int {\mathrm{cosec}}^2\mathrm{x}\mathrm{dx}=-\mathrm{cotx}+\mathrm{c}\right] \\ =-\left[cot\frac{\mathrm{\pi }}{2}-cot\frac{\mathrm{\pi }}{4}\right] \\ =-\left[0-1\right] \\ =1 \end{gathered}$$

Thus, option (B) is the correct answer.