Integral Calculus Question 333
Question: $ \int_{{}}^{{}}{\sqrt{1+\cos x}\ dx} $ equals
[RPET 1996]
Options:
A) $ 2\sqrt{2}\sin \frac{x}{2}+c $
B) $ -2\sqrt{2}\sin \frac{x}{2}+c $
C) $ -2\sqrt{2}\cos \frac{x}{2}+c $
D) $ 2\sqrt{2}\cos \frac{x}{2}+c $
Show Answer
Answer:
Correct Answer: A
Solution:
$ I=\int_{{}}^{{}}{\sqrt{1+\cos x},dx}=\int_{{}}^{{}}{\sqrt{2{{\cos }^{2}}(x/2)}dx} $ $ I=\sqrt{2}\int_{{}}^{{}}{\cos (x/2,)dx}=2\sqrt{2}\sin (x/2)+c $ .
BETA
