Integral Calculus Question 45
Question: $ \int_{{}}^{{}}{\sqrt{1+\sin x}\ dx=} $
[MP PET 1995]
Options:
A) $ \frac{1}{2}( \sin \frac{x}{2}+\cos \frac{x}{2} )+c $
B) $ \frac{1}{2}( \sin \frac{x}{2}-\cos \frac{x}{2} )+c $
C) $ 2\sqrt{1+\sin x}+c $
D) $ -2\sqrt{1-\sin x}+c $
Show Answer
Answer:
Correct Answer: D
Solution:
$ \int_{{}}^{{}}{\sqrt{1+\sin x},dx}=\int_{{}}^{{}}{\sqrt{{{( \sin \frac{x}{2}+\cos \frac{x}{2} )}^{2}}}}dx $ $ =\int_{{}}^{{}}{\sin \frac{x}{2},dx}+\int_{{}}^{{}}{\cos \frac{x}{2},dx}=-2\cos \frac{x}{2}+2\sin \frac{x}{2}+c $ $ =-2( \cos \frac{x}{2}-\sin \frac{x}{2} )+c=-2\sqrt{(1-\sin x)}+c. $
BETA
