Question

The value of ${\int }_{\pi /4}^{\pi /2}{e}^x(\log \sin x+\cot x)dx$ is

• A. $e^{\pi /4}\log 2$
• B. $-e^{\pi /4}\log 2$
• C. $\frac{1}{2}{e}^{\pi /4}\log 2$
• D. $-\frac{1}{2}{e}^{\pi /4}\log 2$

Answer 1

The correct option is C $\frac{1}{2}{e}^{\pi /4}\log 2$

Let $I={\int }_{\pi /4}^{\pi /2}{e}^x(\log \sin x+\cot x)dx$
$\Rightarrow I={\int }_{\pi /4}^{\pi /2}{e}^x\log \sin xdx+{\int }_{\pi /4}^{\pi /2}{e}^x\cot xdx$
$={\int }_{\pi /4}^{\pi /2}{e}^x\log \sin xdx+{\left[e^x\log \sin x\right]}_{\pi /4}^{\pi /2}-{\int }_{\pi /4}^{\pi /2}{e}^x\log \sin xdx$
$=e^{\pi /2}\log \sin \frac{\pi }{2}-e^{\pi /4}\log \sin \frac{\pi }{4}$
$=-e^{\pi /4}\log \left(\frac{1}{\sqrt{2}}\right)$
$=\frac{1}{2}{e}^{\pi /4}\log 2$