Integral Calculus Question 279
Question: $ \int_{{}}^{{}}{\frac{1}{\sqrt{1+\cos x}}\ dx=} $
Options:
A) $ \sqrt{2}\log ( \sec \frac{x}{2}+\tan \frac{x}{2} )+K $
B) $ \frac{1}{\sqrt{2}}\log ( \sec \frac{x}{2}+\tan \frac{x}{2} )+K $
C) $ \log ( \sec \frac{x}{2}+\tan \frac{x}{2} )+K $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
$ \int_{{}}^{{}}{\frac{1}{\sqrt{1+\cos x}}},dx=\int_{{}}^{{}}{\frac{dx}{\sqrt{2{{\cos }^{2}}(x/2)}}}=\frac{1}{\sqrt{2}}\int_{{}}^{{}}{\sec \frac{x}{2},dx} $ $ =\frac{1}{\sqrt{2}}{ \log ( \sec \frac{x}{2}+\tan \frac{x}{2} ) }.\frac{1}{1/2}=\sqrt{2}\log ( \sec \frac{x}{2}+\tan \frac{x}{2} )+K $ .
BETA
