Integral Calculus Question 170
Question: $ \int_{{}}^{{}}{{{\tan }^{-1}}\sqrt{\frac{1-\cos 2x}{1+\cos 2x}}}\ dx= $
Options:
A) $ 2x^{2}+c $
B) $ x^{2}+c $
C) $ \frac{x^{2}}{2}+c $
D) $ 2x+c $
Show Answer
Answer:
Correct Answer: C
Solution:
$ \int_{{}}^{{}}{{{\tan }^{-1}}\sqrt{\frac{1-\cos 2x}{1+\cos 2x}},dx=\int_{{}}^{{}}{{{\tan }^{-1}}\tan x,dx}} $ $ =\int_{{}}^{{}}{x,dx=\frac{x^{2}}{2}}+c. $
BETA
