Integral Calculus Question 353
Question: $ \int{{{\cos }^{-1}}( \frac{1}{x} )dx} $
[RPET 2002]
Options:
A) $ x{{\sec }^{-1}}x+{{\cosh }^{-1}}x+C $
B) $ x{{\sec }^{-1}}x-{{\cosh }^{-1}}x+C $
C) $ x{{\sec }^{-1}}x-{{\sin }^{-1}}x+C $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
$ I=\int{{{\cos }^{-1}}( \frac{1}{x} ),dx} $ $ =\int{{{\sec }^{-1}}x.1dx} $ $ ={{\sec }^{-1}}x\int{dx-\int{[ \frac{d}{dx}{{\sec }^{-1}}x\int{dx,} ]},dx} $ $ =x{{\sec }^{-1}}x-\int{\frac{1}{x\sqrt{x^{2}-1}}x.dx} $ $ =x{{\sec }^{-1}}x-\int{\frac{1}{\sqrt{x^{2}-1}}dx} $ $ =x{{\sec }^{-1}}x-{{\cosh }^{-1}}x+c $ .
BETA
