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:
$\endgroup$ 6(i) $\frac{\sinh^2(x)}{2}+C$, (ii) $\frac{\cosh^2(x)}{2} +C$, and (iii) $\frac{1}{4}\cosh(2x) +C$
2 Answers
$\begingroup$The indefinite integral is defined only up to an additive constant. So all your answers are correct! You may verify this by differentiating your answer and you should get the original function back.
Note that $\cosh (2x) = 2\cosh^2(x)-1 = 2\sinh^2(x)+1$
$\endgroup$ $\begingroup$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: $$\int \sinh x \cosh x \ \mathrm{d}x = \int u \ \mathrm{d}u = \frac{u^2}{2} + c= \frac{\sinh^2 x}{2} + c.$$
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