Proving that the antiderivative of $\operatorname{sech}^2$ is $\operatorname{tanh}$

Question

I just had this question on the exams. I tried many ways but didn't figure it out. Can someone help?

Thanks in advance.

Answer 1

The simplest answer could use substitution $t=\tanh x \Rightarrow \cosh^2 x=\frac{1}{1-t^2}; dx=\frac{dt}{1-t^2}\Rightarrow$ $$ \int\frac{dx}{\cosh^2x}=\int dt= t+C=\tanh x +C. $$

However by careful reading one notes that the derivation just replaces the preknowledge of the derivative of $\tanh x$ with in fact equivalent preknowledge of the derivative of ${\rm arctanh}\, x$.

Therefore it is preferrable to choose another way which uses only derivatives of logarithmic and power functions: $$ \int\frac{dx}{\cosh^2x}=\int\frac{4dx}{(e^x+e^{-x})^2}\stackrel{e^x\mapsto u}= \int\frac{4}{(u+u^{-1})^2}\frac{du}{u}=\int\frac{4udu}{(u^2+1)^2}\\ =-\frac{2}{u^2+1}+(1+C)=\frac{u^2-1}{u^2+1}+C=\frac{u-u^{-1}}{u+u^{-1}}+C \stackrel{u\mapsto e^x}=\tanh x +C. $$