What is the integral of $\cosh^3(x)$? And how exactly can I calculate it? I've tried setting $\cosh^3(x)=(\frac{e^x+e^{-x}}{2})^3$ but all I get in the end is one long fraction.
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$\begingroup$Hint: $\cosh^3x=\cosh x\cdot\cosh^2x=\sinh'x\cdot\big(1+\sinh^2x\big)$.
$\endgroup$ 0 $\begingroup$Hint this is better to integrate $$\frac{e^{-3 x}}{8}+\frac{3 e^{-x}}{8}+\frac{3 e^x}{8}+\frac{e^{3 x}}{8}$$ integrating this we get $$-\frac{e^{-3x}}{24}-\frac{3}{8}e^{-x}+\frac{3}{8}e^{x}+\frac{e^{3x}}{24}+C$$
$\endgroup$ 4 $\begingroup$$$\int \cosh^3(x)dx=\int (\frac{e^x+e^{-x}}{2})^3dx=\frac{1}{8}\int (e^{3x}+3e^{x}+3e^{-x}+e^{-3x})dx$$ $$=\frac{1}{8} (\frac{1}{3}e^{3x}+3e^{x}-3e^{-x}-\frac{1}{3}e^{-3x})+C$$ $$=\frac{2}{24}\sinh(3x)+\frac{3}{4}\sinh(x)+C$$
$\endgroup$ 1 $\begingroup$\begin{align} \int\cosh^3x dx & = \int\left(\sinh^2x +1\right)\cosh x dx \\ & = \frac{1}{3}\int3\sinh^2x\cosh x dx +\int \cosh x dx \\ & = \frac{1}{3}\sinh^3 x + \sinh x + C\\ \end{align}
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