How can I prove that
∫sech(x) dx=sin−1(tanh(x))+c?
I don’t know how to prove this identity. Any help?
I tried to multiply by cosh(x)cosh(x), and everything is okay, but at last I didn’t get the same answer.
Answer
Since you know the result, why not just differentiate it:
(arcsin(tanhx))′=(tanhx)′×1√1−(tanhx)2=(sechx)2×1√(sechx)2=sechx.
I don’t know how to prove this identity. Any help?
This proves that
∫sechxdx=arcsin(tanhx)
up to a constant.
Edit: you might want to write
∫sechxdx=∫(sechx)2×1√(sechx)2dx=∫(tanhx)′×1√1−(tanhx)2dx=arcsin(tanhx)+C.
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