Evaluate the indefinite integral integrate of arcsec(x)dx

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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`

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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

 

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