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

MCQ

$\int_{}^{} {{e^{x/2}}\sin \left( {\frac{x}{2} + \frac{\pi }{4}} \right)\;dx = } $

A
${e^{x/2}}\cos \frac{x}{2} + c$
B
$\sqrt 2 {e^{x/2}}\cos \frac{x}{2} + c$
C
${e^{x/2}}\sin \frac{x}{2} + c$
✓
$\sqrt 2 {e^{x/2}}\sin \frac{x}{2} + c$

✓

Answer

Correct option: D.

$\sqrt 2 {e^{x/2}}\sin \frac{x}{2} + c$

d
(d) Let $I = \int_{}^{} {{e^{x2}}\sin \left( {\frac{x}{2} + \frac{\pi }{4}} \right)dx} $
$ = 2\sin \left( {\frac{x}{2} + \frac{\pi }{4}} \right)\,{e^{x2}} - \int_{}^{} {\cos \left( {\frac{x}{2} + \frac{\pi }{4}} \right)\frac{1}{2}2{e^{x2}}dx + c} $
$ = 2\sin \left( {\frac{x}{2} + \frac{\pi }{4}} \right){e^{x2}} - 2{e^{x2}}\cos \,\left( {\frac{x}{2} + \frac{\pi }{4}} \right) - \int_{}^{} {\sin \left( {\frac{x}{2} + \frac{\pi }{4}} \right)\frac{1}{2}2{e^{x2}}} $
Therefore, $2I = 2{e^{x2}}\left\{ {\sin \left( {\frac{x}{2} + \frac{\pi }{4}} \right) - \cos \left( {\frac{x}{2} + \frac{\pi }{4}} \right)} \right\}$
$ \Rightarrow I = {e^{x2}}\left\{ {\sin \left( {\frac{x}{2} + \frac{\pi }{4}} \right) - \cos \left( {\frac{x}{2} + \frac{\pi }{4}} \right)} \right\}$
$ = \sqrt 2 {e^{x2}}\left( {\sin \frac{x}{2}} \right) = \sqrt 2 {e^{x2}}\sin \frac{x}{2} + c.$
Trick : By inspection,
$\frac{d}{{dx}}\left\{ {\sqrt 2 {e^{x2}}\sin \frac{x}{2} + c} \right\} = \sqrt 2 \left[ {\frac{1}{2}{e^{x2}}\cos \frac{x}{2} + \frac{1}{2}{e^{x2}}\sin \frac{x}{2}} \right]$
$ = {e^{x2}}\left[ {\frac{1}{{\sqrt 2 }}\cos \frac{x}{2} + \frac{1}{{\sqrt 2 }}\sin \frac{x}{2}} \right] = {e^{x2}}\sin \left( {\frac{x}{2} + \frac{\pi }{4}} \right)$.

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