Integral Calculus Question 426
Question: $ \int_{{}}^{{}}{\frac{1}{{{\cos }^{-1}}x.\sqrt{1-x^{2}}}dx=} $
Options:
A) $ \log ({{\cos }^{-1}}x)+c $
B) $ -\log ({{\cos }^{-1}}x)+c $
C) $ -\frac{1}{2{{({{\cos }^{-1}}x)}^{2}}}+c $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
Put $ {{\cos }^{-1}}x=t\Rightarrow -\frac{1}{\sqrt{1-x^{2}}},dx=dt, $ then $ \int_{{}}^{{}}{\frac{1}{{{\cos }^{-1}}\sqrt{1-x^{2}}},dx=-\int_{{}}^{{}}{\frac{1}{t},dt}}=-\log t+c=\log \frac{1}{t}+c $ $ =-\log ({{\cos }^{-1}}x)+c. $
BETA
