Integral Calculus Question 302
Question: $ \int_{{}}^{{}}{\frac{1}{\sqrt{1+\sin x}}dx}= $
[RPET 1996]
Options:
A) $ 2\sqrt{2}\log \tan ( \frac{\pi }{8}+\frac{x}{4} )+c $
B) $ \frac{1}{\sqrt{2}}\log \tan ( \frac{\pi }{8}+\frac{x}{4} )+c $
C) $ \sqrt{2}\log \tan ( \frac{\pi }{8}+\frac{x}{4} )+c $
D) $ \frac{1}{2\sqrt{2}}\log \tan ( \frac{\pi }{8}+\frac{x}{4} )+c $
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Answer:
Correct Answer: C
Solution:
$ \int_{{}}^{{}}{\frac{1}{\sqrt{1+\sin x}}},dx=\int_{{}}^{{}}{\frac{1}{\sqrt{2}\sin ( \frac{\pi }{4}+\frac{x}{2} )}},dx $ $ =\frac{1}{\sqrt{2}}\int_{{}}^{{}}{cosec,( \frac{x}{2}+\frac{\pi }{4} )},dx=\sqrt{2}\log \tan ( \frac{\pi }{8}+\frac{x}{4} )+c. $
BETA
