• October 28th 2006, 11:11 AM
Yogi_Bear_79
Evaluate: $\int sinh^6 x\ cosh\ xdx$

$=\frac{1}{2}x^2\ cos(h) \sin(h^6)+C$
• October 28th 2006, 12:07 PM
topsquark
Quote:

Originally Posted by Yogi_Bear_79
Evaluate: $\int sinh^6 x\ cosh\ xdx$

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

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

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

-Dan
• October 28th 2006, 02:28 PM
CaptainBlack
Quote:

Originally Posted by Yogi_Bear_79
Evaluate: $\int sinh^6 x\ cosh\ xdx$

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

$
\int \sinh^6 (x)\ \cosh(x) dx
$

That is the functions are hyperbolic not trignometric, and so the h is
part of the function name and not a variable.

RonL
• October 28th 2006, 02:47 PM
Yogi_Bear_79
well that does help alot. But leads to another question.

is this the same as the orignal equation?
$\int sinh (x)^6 cosh(x)$
• October 28th 2006, 03:03 PM
CaptainBlack
Quote:

Originally Posted by Yogi_Bear_79
well that does help alot. But leads to another question.

is this the same as the orignal equation?
$\int sinh (x)^6 cosh(x)$

You need to be carefull where the power goes. There is a convention that $\sinh^n(x)$ actually means $[\sinh(x)]^n$ (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:

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

RonL