# Thread: Integration with Hyperbolic Sine

1. ## Integration with Hyperbolic Sine

These freak me out. Haha.

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

What do I do?

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

Then its str8forward right

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

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

4. 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 $\displaystyle e$, 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.

5. 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 $\displaystyle cothx$.

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

So check my work:

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

I must still be doing something wrong though, because the book says the answer should be $\displaystyle cothx$.
The red is wrong.
$\displaystyle ln(2)$ is a constant.
The derivative of the constant is zero.

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

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

9. Originally Posted by xfriendsonfirex
These freak me out. Haha.

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

What do I do?
If you're trying to differentiate it...

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

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

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

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

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

$\displaystyle = \coth{x}$.

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

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

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

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

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

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

$\displaystyle = \coth{x}$.

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.