Integral $\int \operatorname{sech}^4 x \, dx$

Question

How we can solve this?$\newcommand{\sech}{\operatorname{sech}}$ $$ \int \sech^4 x \, dx. $$ I know we can solve the simple case $$ \int \sech \, dx=\int\frac{dx}{\cosh x}=\int\frac{dx\cosh x}{\cosh ^2x}=\int\frac{d(\sinh x)}{1+\sinh^2 x}=\int \frac{du}{1+u^2}=\tan^{-1}\sinh x+C. $$ I am stuck with the $\sech^4$ though. Thank you

Answer 1

Note that $$ \int \DeclareMathOperator{sech}{sech}{\sech}^4x\,dx=\int{\sech}^2{x}\cdot(1-\tanh^2x)\,dx $$ Letting $u=\tanh x$ gives $du={\sech}^2x$ so $$ \int{\sech}^4x\,dx=\int(1-u^2)\,du=u-\frac{u^3}{3}+C=\tanh x-\frac{1}{3}\tanh^3x+C $$

Answer 2

Since \begin{align} \partial_{x} \left[ \tanh^{m}(ax) \right] = am\ sech^{2}(ax) \ \tanh^{m-1}(ax) \end{align} then \begin{align} sech^{4}(ax) &= sech^{2}(ax) (1 - \tanh^{2}(ax) ) \\ &= \partial_{x} \left[ \frac{1}{a} \tanh(ax) \right] - \partial_{x} \left[ \frac{1}{3a} \tanh^{3}(ax) \right] \\ &= \partial_{x} \left[ \frac{1}{a} \tanh(ax) - \frac{1}{3a} \tanh^{3}(ax) \right]. \end{align} Upon integration of both sides the result is \begin{align} \int sech^{4}(ax) \ dx = \frac{1}{a} \tanh(ax) - \frac{1}{3a} \tanh^{3}(ax). \end{align}