# Thread: integrals of inverse trig

1. ## integrals of inverse trig

show that sec-1=cos-1(1/x)

2. ## Re: integrals of inverse trig

Originally Posted by Wattssan
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.

3. ## Re: integrals of inverse trig

Originally Posted by Wattssan
show that sec-1(?)=cos-1(1/x)
There is something missing in the post.