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

Answer 1

Samantha Adkison

#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#

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Answer 2

Luke Alvarez

To integrate ( \text{sech}(x) (\tanh(x) - \text{sech}(x)) ) with respect to ( x ), you can use substitution. Let ( u = \tanh(x) - \text{sech}(x) ). Then ( du = (\text{sech}^2(x) - \text{sech}(x)\tanh(x))dx ). Now, rewrite the integral in terms of ( u ) and ( du ). This will result in a simpler integral that you can solve.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.