Integrate $\int \cosh^4(7x) dx$

Question

Can you help me with solving this integral $$\int \cosh^4(7x) \, dx \text{ ?}$$ I tried use subs but i got $$\cosh^4(t)=m$$ $$t = \pm\operatorname{arccosh}^{-1}(m^{1/4})$$ $$dt = \pm\operatorname{arccosh}(m^{1/4})\frac14m^{{-3/4}} \, dm$$

What now?

Answer 1

Substitute $\cosh (7x)=\large \frac{e^{7x}+e^{-7x}}{2}$, $$\int \cosh^4 (7x)\ dx=\int \left(\frac{e^{7x}+e^{-7x}}{2}\right)^4\ dx$$ $$=\int \frac{e^{28x}+e^{-28x}+e^{14x}+e^{-14x}+6}{16}\ dx$$ $$=\frac{1}{16}\left(\frac{e^{28x}}{28}-\frac{e^{-28x}}{28}+\frac{e^{14x}}{14}-\frac{e^{-14x}}{14}+6x\right)+C$$ $$=\frac{1}{16}\left(\frac{\sinh (28x)}{28}+\frac{\sinh (14x)}{14}+6x\right)+C$$

Answer 2

Hint: $$\cosh^2x=\frac{1+\cosh2x}{2}$$

Answer 3

HINT...use the "double angle" identity $$\cosh^2A=\frac 12(1+\cosh 2A) $$ twice to get the integrand in terms of multiple angles

Answer 4

Hint:

As for trigonometric functions, the solution is to linearise first: \begin{align*}\cosh^4u&=\frac1{16}(\mathrm e^u+\mathrm e^{-u})^4=\frac1{16}(\mathrm e^{4u}+4\mathrm e^{2u}+6+4\mathrm e^{-2u}+\mathrm e^{-4u})\\ &=\frac18\cosh4u+\frac12\cosh2u+\frac38. \end{align*}