Solveeit Logo
Login

Question

Question: \(\int _ { 0 } ^ { \pi / 2 } e ^ { x } \sin x d x =\)...

∫0π/2exsin⁡xdx=\int _ { 0 } ^ { \pi / 2 } e ^ { x } \sin x d x =

A

12(eπ/2−1)\frac { 1 } { 2 } \left( e ^ { \pi / 2 } - 1 \right)

B

12(eπ/2+1)\frac { 1 } { 2 } \left( e ^ { \pi / 2 } + 1 \right)

C

12(1−eπ/2)\frac { 1 } { 2 } \left( 1 - e ^ { \pi / 2 } \right)

D

2(eπ/2+1)2 \left( e ^ { \pi / 2 } + 1 \right)

Answer

12(eπ/2+1)\frac { 1 } { 2 } \left( e ^ { \pi / 2 } + 1 \right)

Explanation

Solution

Let I=∫0π/2exsin⁡xdxI = \int _ { 0 } ^ { \pi / 2 } e ^ { x } \sin x d x

= −[excos⁡x]0π/2+∫0π/2excos⁡xdx- \left[ e ^ { x } \cos x \right] _ { 0 } ^ { \pi / 2 } + \int _ { 0 } ^ { \pi / 2 } e ^ { x } \cos x d x

=−[excos⁡x]0π/2+[exsin⁡x]0π/2−∫0π/2exsin⁡xdx= - \left[ e ^ { x } \cos x \right] _ { 0 } ^ { \pi / 2 } + \left[ e ^ { x } \sin x \right] _ { 0 } ^ { \pi / 2 } - \int _ { 0 } ^ { \pi / 2 } e ^ { x } \sin x d x

2I=[ex(sin⁡x−cos⁡x)]0π/2=(eπ/2+1)2 I = \left[ e ^ { x } ( \sin x - \cos x ) \right] _ { 0 } ^ { \pi / 2 } = \left( e ^ { \pi / 2 } + 1 \right)

Hence ∫0π/2exsin⁡xdx=12(eπ/2+1)\int _ { 0 } ^ { \pi / 2 } e ^ { x } \sin x d x = \frac { 1 } { 2 } \left( e ^ { \pi / 2 } + 1 \right)