How do you integrate #xsec^2x#?

Question

I tried doing by substitution where U = x, du= 1, dv = sec^2x, v = tanx.

Then integrate UV- integral of vdu. Then another substitution in the following equation, but I got integral of xsec^2x = x tanx- integral of xsec^2) which is wrong.

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Answer 1 · 2018-04-01T08:01:24

The answer is #=xtanx+ln(|cosx|)+C#

#" Reminder "#

#inttanxdx=-ln(|cosx|)+C#

#intuv'=uv-intu'v#

#u=x#, #=>#, #u'=1#

#v'=sec^2x#, #=>#, #v=tanx#

Therefore,

#intxsec^2xdx=xtanx-inttanxdx#

#=xtanx+ln(|cosx|)+C#