# Integrating $\sinh(x)\cosh(x)$

So I am very new to integration. I have to find the integral of $\sinh(x)\cosh(x)$

I have tried different ways:

(i) let $u = \sinh(x)$, (ii) let $u= \cosh(x)$, and (iii) using the identity $\sinh(2x) = 2 \sinh(x)\cosh(x)$

However, all of these result in different answers. In particular the answers are:

(i) $\frac{\sinh^2(x)}{2}+C$, (ii) $\frac{\cosh^2(x)}{2} +C$, and (iii) $\frac{1}{4}\cosh(2x) +C$

• The derivative of $\sinh x$ with respect to $x$ is $\cosh x$. The substitution $u=\sinh x$ is then the appropriate route to take. Sep 13 '14 at 2:00
• so, do you mean that the other answers are incorrect? Sep 13 '14 at 2:02
• Fun fact: $\cosh^2(x)-\sinh^2(x)=1$ Sep 13 '14 at 2:02
• @Aaron Well, not necessarily. The substitution $u=\cosh x$ would work as well, since the derivative of such is $\sinh x$. I would use either i or ii, it does not really matter. Sep 13 '14 at 2:04
• Guys, his answers all differ by a constant. That is all there is to it. Sep 13 '14 at 2:06

Note that $\cosh (2x) = 2\cosh^2(x)-1 = 2\sinh^2(x)+1$
Remember that, by definition, we have: $$\sinh x = \frac{e^x - e^{-x}}{2} \quad \mbox{and} \quad \cosh x = \frac{e^x + e^{-x}}{2}$$ It is an easy exercise to check that $\frac{\mathrm{d}}{\mathrm{d}x}\sinh x = \cosh x$ and $\frac{\mathrm{d}}{\mathrm{d}x}\cosh x = \sinh x$. So, making $u = \sinh x$, we have $\mathrm{d}u = \cosh x \ \mathrm{d}x$, and hence: $$\int \sinh x \cosh x \ \mathrm{d}x = \int u \ \mathrm{d}u = \frac{u^2}{2} + c= \frac{\sinh^2 x}{2} + c.$$