show that sec^{-1}=cos^{-1}(1/x)
That question does not make sense. You cannot apply an inverse trigonometric function to an equals sign.
Do you mean $\sec^{-1} x = \cos^{-1}\left( \dfrac{1}{x} \right)$?
$y = \sec^{-1} x \Longrightarrow x = \sec y \Longrightarrow \dfrac{1}{x} = \cos y \Longrightarrow y = \cos^{-1}\left( \dfrac{1}{x}\right)$
This is not a question of integration, so I do not understand the title of your post.