Integral: Verify Anwser Please

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October 28th 2006, 10:11 AM
Yogi_Bear_79

Integral: Verify Anwser Please

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

$=\frac{1}{2}x^2\ cos(h) \sin(h^6)+C$
October 28th 2006, 11:07 AM
topsquark

Quote:

Originally Posted by Yogi_Bear_79

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

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

There seems to be a problem with your given answer.

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, 01:28 PM
CaptainBlack

Quote:

Originally Posted by Yogi_Bear_79

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

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

Your integral is:

$<br /> \int \sinh^6 (x)\ \cosh(x) dx<br />$

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, 01:47 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, 02:03 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