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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) $\frac{1}{4}\cosh(2x) +C$

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    $\begingroup$ The derivative of $\sinh x$ with respect to $x$ is $\cosh x$. The substitution $u=\sinh x$ is then the appropriate route to take. $\endgroup$
    – Gahawar
    Sep 13 '14 at 2:00
  • $\begingroup$ so, do you mean that the other answers are incorrect? $\endgroup$
    – Aaron
    Sep 13 '14 at 2:02
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    $\begingroup$ Fun fact: $\cosh^2(x)-\sinh^2(x)=1$ $\endgroup$
    – fixedp
    Sep 13 '14 at 2:02
  • $\begingroup$ @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. $\endgroup$
    – Gahawar
    Sep 13 '14 at 2:04
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    $\begingroup$ Guys, his answers all differ by a constant. That is all there is to it. $\endgroup$
    – fixedp
    Sep 13 '14 at 2:06
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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 $\cosh (2x) = 2\cosh^2(x)-1 = 2\sinh^2(x)+1$

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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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