Integral Calculus Question 438
Question: $ \int_{{}}^{{}}{\frac{dx}{x\sqrt{1-{{(\log x)}^{2}}}}=} $
Options:
A) $ {{\cos }^{-1}}(\log x)+c $
B) $ x\log (1-x^{2})+c $
C) $ {{\sin }^{-1}}(\log x)+c $
D) $ \frac{1}{2}{{\cos }^{-1}}(\log x)+c $
Show Answer
Answer:
Correct Answer: C
Solution:
Put $ \log x=t\Rightarrow \frac{1}{x},dx=dt, $ then it reduces to $ \int_{{}}^{{}}{\frac{dt}{\sqrt{1-t^{2}}}={{\sin }^{-1}}t={{\sin }^{-1}}(\log x)+c.} $
BETA
