How do you integrate #sechx (tanhx-sechx) dx#?

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Answer 1 · 2016-04-21T11:13:58

#sechx-tanhx+C#

If we distribute the #sechx# term, we get

#=int(sechxtanhx-sech^2x)dx#

Splitting this into two integrals:

#=intsechxtanhxdx-intsech^2xdx#

Recall that #(sechx)'=sechxtanhx# and #(tanhx)'=sech^2x#.

Thus, the integral gives us:

#=sechx-tanhx+C#