Integration with Hyperbolic Sine

**xfriendsonfirex** · February 15th 2010, 01:37 PM

These freak me out. Haha.

$f(x)=ln(sinhx)$

What do I do?

**vince** · February 15th 2010, 01:39 PM

express $sinh$ as $\frac{e^x-e^{-x}} {2}$

Then its str8forward right

**xfriendsonfirex** · February 15th 2010, 01:44 PM

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

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

**vince** · February 15th 2010, 02:11 PM

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.

**xfriendsonfirex** · February 15th 2010, 02:20 PM

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 $cothx$ .

**General** · February 15th 2010, 02:22 PM

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.

**xfriendsonfirex** · February 15th 2010, 02:33 PM

But even with that corrected, I still don't see how I get $cothx$ .

**General** · February 15th 2010, 03:07 PM

$\frac{d}{dx} ln(f(x)) = \frac{f'(x)}{f(x)}$ . for $f>0$ .

**Prove It** · February 15th 2010, 04:26 PM

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 $y = \ln{u}$ .

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

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

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

$= \coth{x}$ .

**vince** · February 15th 2010, 04:55 PM

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.

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