These freak me out. Haha.

$\displaystyle f(x)=ln(sinhx)$

What do I do?

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- Feb 15th 2010, 01:37 PMxfriendsonfirexIntegration with Hyperbolic Sine
These freak me out. Haha.

$\displaystyle f(x)=ln(sinhx)$

What do I do? - Feb 15th 2010, 01:39 PMvince
express $\displaystyle sinh$ as $\displaystyle \frac{e^x-e^{-x}} {2}$

Then its str8forward right - Feb 15th 2010, 01:44 PMxfriendsonfirex
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, 02:11 PMvince

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

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

(i)$\displaystyle ln(\frac{a}{b}) = ln(a)-ln(b)$

(ii)$\displaystyle ln(e^{(x)}) = x$ or $\displaystyle e^{aln(x)} = x^a$

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

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

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

The last equality is now your integrand. Uber-easy expression to integrate. - Feb 15th 2010, 02:20 PMxfriendsonfirex
And actually, I don't even need to integrate, just differentiate.

So check my work:

$\displaystyle f(x)=2x-ln(2)$

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

I must still be doing something wrong though, because the book says the answer should be $\displaystyle cothx$. - Feb 15th 2010, 02:22 PMGeneral
- Feb 15th 2010, 02:33 PMxfriendsonfirex
But even with that corrected, I still don't see how I get $\displaystyle cothx$.

- Feb 15th 2010, 03:07 PMGeneral
$\displaystyle \frac{d}{dx} ln(f(x)) = \frac{f'(x)}{f(x)}$. for $\displaystyle f>0$.

- Feb 15th 2010, 04:26 PMProve It
If you're trying to differentiate it...

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

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

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

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

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

$\displaystyle = \coth{x}$. - Feb 15th 2010, 04:55 PMvince

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:

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

What i thought i saw when i did that was(the following is true):

$\displaystyle ln(e^x)-ln(e^{-x})-ln(2)$ = $\displaystyle 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.