Integral Calculus Question 498
Question: $ \int_{{}}^{{}}{{e^{x/2}}\sin ( \frac{x}{2}+\frac{\pi }{4} )\ dx=} $
[Roorkee 1982]
Options:
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 $
D) $ \sqrt{2}{e^{x/2}}\sin \frac{x}{2}+c $
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Answer:
Correct Answer: D
Solution:
Let $ I=\int_{{}}^{{}}{{e^{x/2}}\sin ( \frac{x}{2}+\frac{\pi }{4} )dx} $
$ =2\sin ( \frac{x}{2}+\frac{\pi }{4} ),{e^{x/2}}-\int_{{}}^{{}}{\cos ( \frac{x}{2}+\frac{\pi }{4} )\frac{1}{2}2{e^{x/2}}dx+c} $ $ =2\sin ( \frac{x}{2}+\frac{\pi }{4} ){e^{x/2}}-2{e^{x/2}}\cos ,( \frac{x}{2}+\frac{\pi }{4} )-\int_{{}}^{{}}{\sin ( \frac{x}{2}+\frac{\pi }{4} )\frac{1}{2}2{e^{x/2}}} $
Therefore, $ 2I=2{e^{x/2}}{ \sin ( \frac{x}{2}+\frac{\pi }{4} )-\cos ( \frac{x}{2}+\frac{\pi }{4} ) } $
$ \Rightarrow I={e^{x/2}}{ \sin ( \frac{x}{2}+\frac{\pi }{4} )-\cos ( \frac{x}{2}+\frac{\pi }{4} ) } $
$ =\sqrt{2}{e^{x/2}}( \sin \frac{x}{2} )=\sqrt{2}{e^{x/2}}\sin \frac{x}{2}+c. $
Trick : By inspection,
$ \frac{d}{dx}{ \sqrt{2}{e^{x/2}}\sin \frac{x}{2}+c }=\sqrt{2}[ \frac{1}{2}{e^{x/2}}\cos \frac{x}{2}+\frac{1}{2}{e^{x/2}}\sin \frac{x}{2} ] $
$ ={e^{x/2}}[ \frac{1}{\sqrt{2}}\cos \frac{x}{2}+\frac{1}{\sqrt{2}}\sin \frac{x}{2} ]={e^{x/2}}\sin ( \frac{x}{2}+\frac{\pi }{4} ) $ .
BETA
