1. ## Integral: Verify Anwser Please

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

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

2. 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

3. 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

4. well that does help alot. But leads to another question.

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

5. 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