Explanation

(b)
$\int_{}^{}{\cos^{- 1}{\left( \frac{1}{x} \right)dx}} = \int_{}^{}{\sec^{- 1}{x \bullet 1\ dx}}$
$= \sec^{- 1}{x\int_{}^{}{dx - \int_{}^{}{\left\lbrack \frac{d}{dx}\sec^{- 1}{x\int_{}^{}{dx}} \right\rbrack dx}}}$
$= x\sec^{- 1}{x - \int_{}^{}{\frac{1}{x\sqrt{x^{2} - 1}}x\ dx}}$
$= x\sec^{- 1}{x -}\int_{}^{}{\frac{1}{\sqrt{x^{2} - 1}}\ dx}$
$= x\sec^{- 1}{x -}\cosh^{- 1}{x + c}$