Evaluate the indefinite integral integrate of arcsec(x)dx

1 Answer | Add Yours

mathsworkmusic's profile pic

mathsworkmusic | (Level 2) Educator

Posted on

You need to integrate by parts

`int u (dv)/dx dx = uv - int v (du)/dx dx `

Let `v = x implies (dv)/dx = 1`

And `u = arcsecx` `implies (du)/dx = 1/(|x| sqrt(x^2 -1))` (see wikipedia page below)

Then `int arcsecx = xarcsecx - int x/(|x|sqrt(x^2-1)) dx`

Using integral of one over square root of a binomial and checking the cases `x < -1` and `x>1` (the domain of `x`)

`implies \int arcsecx = xarcsecx - ln|x + sqrt(x^2 -1)| + C` answer



We’ve answered 317,416 questions. We can answer yours, too.

Ask a question