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
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2 Answers
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3 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 $$ \int{\sech}^4x\,dx=\int(1-u^2)\,du=u-\frac{u^3}{3}+C=\tanh x-\frac{1}{3}\tanh^3x+C $$
- 1$\begingroup$ How come mathematica says $$ \int \frac{dx}{\cosh^4 x}=\left(\frac{2\tanh x}{3}+\frac{1}{3}\tanh x \text{sech}^2x\right)? $$ Are they equivalent? It seems the first factor $\tanh x$ and $2/3 \tanh x$ are not. However I do agree with your method. Thanks... $\endgroup$user143444– user1434442014-05-12 20:24:25 +00:00Commented May 12, 2014 at 20:24
- $\begingroup$ Plug $\DeclareMathOperator{sech}{sech}{\sech}^2(x)=1-\tanh^2x$ into your formula. $\endgroup$Brian Fitzpatrick– Brian Fitzpatrick2014-05-12 20:27:25 +00:00Commented May 12, 2014 at 20:27
- $\begingroup$ Thanks a lot. I get it now $\endgroup$user143444– user1434442014-05-12 20:37:42 +00:00Commented May 12, 2014 at 20:37
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1 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}
- $\begingroup$ THanks for the other solution +1 $\endgroup$user143444– user1434442014-05-13 00:16:07 +00:00Commented May 13, 2014 at 0:16