Evaluate: $\displaystyle \int sinh^6 x\ cosh\ xdx$

$\displaystyle =\frac{1}{2}x^2\ cos(h) \sin(h^6)+C$

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- Oct 28th 2006, 10:11 AMYogi_Bear_79Integral: Verify Anwser Please
Evaluate: $\displaystyle \int sinh^6 x\ cosh\ xdx$

$\displaystyle =\frac{1}{2}x^2\ cos(h) \sin(h^6)+C$ - Oct 28th 2006, 11:07 AMtopsquark
There seems to be a problem with your given answer.

Let $\displaystyle y = sinh(x)$, then $\displaystyle dy = cosh(x) dx$

So:

$\displaystyle \int sinh^6(x) \, cosh(x) dx = \int \, y^6 dy$ = $\displaystyle \frac{1}{7}y^7 + C$ = $\displaystyle \frac{1}{7}sinh^7(x) + C$.

-Dan - Oct 28th 2006, 01:28 PMCaptainBlack
- Oct 28th 2006, 01:47 PMYogi_Bear_79
well that does help alot. But leads to another question.

is this the same as the orignal equation?

$\displaystyle \int sinh (x)^6 cosh(x) $ - Oct 28th 2006, 02:03 PMCaptainBlack
You need to be carefull where the power goes. There is a convention that $\displaystyle \sinh^n(x)$ actually means $\displaystyle [\sinh(x)]^n$ (even though this is an abuse of notation). If you move the power to where you have it you will need to add appropriate brackets to make the meaning clear.

So what the original means is:

$\displaystyle \int [\sinh (x)]^6 \cosh(x) $

RonL