_ /4 ^ /2 e^x ( x + x) ,dx = - Vidyadip

MCQ

$\int_{\pi {\rm{/4}}}^{\pi {\rm{/2}}} {{e^x}(\log \sin x + \cot x)\,dx = } $

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

✓

Answer

Correct option: C.

$\frac{1}{2}{e^{\pi /4}}\log 2$

c
(c) Let $I = \int_{\pi /4}^{\pi /2} {{e^x}(\log \sin x + \cot x)dx} $

$I = \int_{\pi /4}^{\pi /2} {{e^x}\log \sin x\,dx + \int_{\pi /4}^{\pi /2} {{e^x}\cot x\,dx} } $

$ = \int_{\pi /4}^{\pi /2} {{e^x}\log \sin xdx + [{e^x}\log \sin x]_{\pi /4}^{\pi /2}} $$ - \int_{\pi /4}^{\pi /2} {{e^x}\log \sin x\,dx} $

$ = {e^{\pi /2}}\log \sin \frac{\pi }{2} - {e^{\pi /4}}\log \sin \frac{\pi }{4} $

$= \frac{1}{2}{e^{\pi /4}}\log 2$.

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