# Integration with Hyperbolic Sine

• Feb 15th 2010, 02:37 PM
xfriendsonfirex
Integration with Hyperbolic Sine
These freak me out. Haha.

$f(x)=ln(sinhx)$

What do I do?
• Feb 15th 2010, 02:39 PM
vince
express $sinh$ as $\frac{e^x-e^{-x}} {2}$

Then its str8forward right
• Feb 15th 2010, 02:44 PM
xfriendsonfirex
If I wasn't horrid at integration it would be. Haha. That definitely helps.

Do I need to use u-substitution here or what?
• Feb 15th 2010, 03:11 PM
vince
Quote:

Originally Posted by xfriendsonfirex
If I wasn't horrid at integration it would be. Haha. That definitely helps.

Do I need to use u-substitution here or what?

you need to brush up on the properties of the logarithm.

two properties used here (applying them to log base $e$, aka, natural logarithm):

(i) $ln(\frac{a}{b}) = ln(a)-ln(b)$
(ii) $ln(e^{(x)}) = x$ or $e^{aln(x)} = x^a$

There are more properties, I HIGHLY RECC' you study them and do problems utilizing them. The premium on knowing the properties of ln, e, as well as those of the core trig functions cannot be overvalued when it comes to calc courses if not the majority of the branches of mathematical sciences!

anyway, we have

$f(x)=ln(sinhx) = ln(\frac{e^x-e^{-x}} {2})$
= $ln(e^x-e^{-x})-ln(2) = ln(\frac{e^x}{e^{-x}})-ln(2) = ln(e^{2x})-ln(2) = 2x-ln(2)$

The last equality is now your integrand. Uber-easy expression to integrate.
• Feb 15th 2010, 03:20 PM
xfriendsonfirex
And actually, I don't even need to integrate, just differentiate.

So check my work:

$f(x)=2x-ln(2)$
$f'(x)=2-\frac{1}{2}=\frac{3}{2}$

I must still be doing something wrong though, because the book says the answer should be $cothx$.
• Feb 15th 2010, 03:22 PM
General
Quote:

Originally Posted by xfriendsonfirex
And actually, I don't even need to integrate, just differentiate.

So check my work:

$f(x)=2x-ln(2)$
$f'(x)=2-{\color{red}\frac{1}{2}}=\frac{3}{2}$

I must still be doing something wrong though, because the book says the answer should be $cothx$.

The red is wrong.
$ln(2)$ is a constant.
The derivative of the constant is zero.
• Feb 15th 2010, 03:33 PM
xfriendsonfirex
But even with that corrected, I still don't see how I get $cothx$.
• Feb 15th 2010, 04:07 PM
General
$\frac{d}{dx} ln(f(x)) = \frac{f'(x)}{f(x)}$. for $f>0$.
• Feb 15th 2010, 05:26 PM
Prove It
Quote:

Originally Posted by xfriendsonfirex
These freak me out. Haha.

$f(x)=ln(sinhx)$

What do I do?

If you're trying to differentiate it...

$y = \ln{\sinh{x}}$.

Let $u = \sinh{x}$ so that $y = \ln{u}$.

$\frac{du}{dx} = \cosh{x}$

$\frac{dy}{du} = \frac{1}{u} = \frac{1}{\sinh{x}}$

$\frac{dy}{dx} = \frac{\cosh{x}}{\sinh{x}}$

$= \coth{x}$.
• Feb 15th 2010, 05:55 PM
vince
Quote:

Originally Posted by Prove It
If you're trying to differentiate it...

$y = \ln{\sinh{x}}$.

Let $u = \sinh{x}$ so that $y = \ln{u}$.

$\frac{du}{dx} = \cosh{x}$

$\frac{dy}{du} = \frac{1}{u} = \frac{1}{\sinh{x}}$

$\frac{dy}{dx} = \frac{\cosh{x}}{\sinh{x}}$

$= \coth{x}$.

sorry just got back, and noticed just now you wanted to differentiate it. as Prove it shows, quite easy. And unfortunately, i just noticed that i made an error above, which is why taking the derivative of the "simplified" form doesnt hold. My error was in claiming:
$ln(e^x-e^{-x})-ln(2) = ln(\frac{e^x}{e^{-x}})-ln(2)$

What i thought i saw when i did that was(the following is true):
$ln(e^x)-ln(e^{-x})-ln(2)$ = $ln(\frac{e^x}{e^{-x}})-ln(2)$

Also, note every other claim is true. just that one equality, while crucial, was wrong.
sorry for the confusion. first time since ive been posting on this forum that ive made mistakes...let alone 2 in one day! when it rains it pours.