Evaluate $\int \cosh^3 (x) \sinh^2 (x )dx $

Question

Evaluate $$\int \cosh^3 (x) \sinh^2 (x )dx $$

So my original thought was to apply the identity that $\sinh^2(x)=\cosh^2(x)-1$. This means that my integral becomes $$\int \cosh^5(x)-\cosh^3(x) dx$$ which is worse to integrate I think. What would be the best approach to tackle this or where can i go from here? Any help would be appreciated.

Answer 1

Let $u=\sinh(x)$. Then $du=\cosh(x)dx$. Hence, the integral is $$\int (u^2+1)u^2du.$$

Answer 2

Hint: Use that $$\cosh^3(x)\cdot \sinh^2(x)=\frac{1}{16} (-2 \cosh (x)+\cosh (3 x)+\cosh (5 x))$$

Answer 3

I guess this could simply be solved using the simple identities: $$2\cosh(x)=e^{x}+e^{-x}, \quad 2\cosh(x)=e^{x}-e^{-x}.$$ For the product $$\cosh^3(x)\sinh ^2(x)=\bigg(\frac{e^{x}+e^{-x}}{2}\bigg)^3\bigg(\frac{e^{x}-e^{-x}}{2}\bigg)^2$$ $$=\frac{1}{32}\big(e^{3x}+e^{-3x}+3(e^{x}-e^{-x})\big)\big(e^{2x}+e^{-2x}-2\big)$$ Expanding the two brackets one obtains: $$\cosh^3(x)\sinh ^2(x)=\frac{1}{32}[(e^{5x}+e^{-5x})+(e^{3x}+e^{-3x})-2(e^{x}+e^{-x})]$$ $$=\frac{1}{32}(2\cosh(5x)+\cosh(3x)-2\cosh (x)).$$ Now they are in a simplified form and easy for integration. On integration of both sides of the above equation we obtain: $$\int\cosh^3(x)\sinh ^2(x)dx=\frac{1}{32}\bigg[ \frac{2\sinh(5x)}{5}+\frac{2\sinh(3x)}{3}-2\sinh(x)\bigg]+\text{constant.}$$ That's it.