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September 10

Integral of inverse cotangent

There's a solution but it seems more complicated than necessary.[1] If you replace the cotangent inverse with tangent, wouldn't the substitution by parts go a lot smoother? u=tan x, so du=sec^2 x dx and the integral of x sec^2 x can be obtained by integration by parts again, dv = sec^2 x dx which when integrated becomes tan x and then I get terms cancelling out and finally

ln |sec x| +C

Where did I go wrong? Imagine Reason (talk) 20:49, 10 September 2016 (UTC)[reply]

I didn't go through any of that but didn't you want to start with x=tan u at the start? So you'd have to substitute in an arctan at the end. Dmcq (talk) 22:41, 10 September 2016 (UTC)[reply]

I confused inverse functions with reciprocals. Yeah, I haven't done this for a while. Thanks. Imagine Reason (talk) 22:51, 10 September 2016 (UTC)[reply]