_0^ /2 e^x x ,dx = - Vidyadip

MCQ

$\int_0^{\pi /2} {{e^x}\sin x\,dx = } $

A
$\frac{1}{2}({e^{\pi /2}} - 1)$
✓
$\frac{1}{2}({e^{\pi /2}} + 1)$
C
$\frac{1}{2}(1 - {e^{\pi /2}})$
D
$2({e^{\pi /2}} + 1)$

✓

Answer

Correct option: B.

$\frac{1}{2}({e^{\pi /2}} + 1)$

b
(b) Let $I = \int_0^{\pi /2} {{e^x}\sin \,x\,\,dx} $

$= - [{e^x}\cos x]_0^{\pi /2} + \int_0^{\pi /2} {{e^x}\cos x\,dx} $

$ = - [{e^x}\cos x]_0^{\pi /2} + [{e^x}\sin x]_0^{\pi /2} - \int_0^{\pi /2} {{e^x}\sin x\,dx} $

$\therefore $$2I = [{e^x}(\sin x - \cos x)]_0^{\pi /2} = ({e^{\pi /2}} + 1)$

Hence $\int_0^{\pi /2} {{e^x}\sin xdx = \frac{1}{2}({e^{\pi /2}} + 1)} $.

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