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integrate hyperbolic sine

bowlbase

New member
Feb 8, 2012
5
integrate Sinh4(x)

I have been struggling with this problem for a week. I know the answer because of wolfram but I cannot see how it gets it. Honestly, I can't even decide what to make my substitution as. I haven't really had problems with any other questions from our homework but this one and its killing me. Thanks for any help or guidance!
 

dwsmith

Well-known member
Feb 1, 2012
1,673
integrate Sinh4(x)

I have been struggling with this problem for a week. I know the answer because of wolfram but I cannot see how it gets it. Honestly, I can't even decide what to make my substitution as. I haven't really had problems with any other questions from our homework but this one and its killing me. Thanks for any help or guidance!
Are you aware that the integral of sinhx is coshx?

Then try this identity

$\sinh^2x = \dfrac{\cosh(2x) - 1}{2}$

and

$\cosh^2x=\dfrac{1+\cosh(2x)}{2}$
 
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bowlbase

New member
Feb 8, 2012
5
(sorry for lack of latex, not sure how to start that on the new forums and I'm a bit out of practice)
Yes, so I originally thought that I should substitute u for sinh. therefore u4 dx, du/dx sinh(x) = cosh(x), then du= cosh(x) dx

so integrate (cosh(x) u4 du)?

how do I get rid of the x?
 

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,403
integrate Sinh4(x)

I have been struggling with this problem for a week. I know the answer because of wolfram but I cannot see how it gets it. Honestly, I can't even decide what to make my substitution as. I haven't really had problems with any other questions from our homework but this one and its killing me. Thanks for any help or guidance!
If you didn't know any of the hyperbolic identities, you could always use its exponential form...

\[ \displaystyle \begin{align*} \sinh^4{x} &= \left[\frac{1}{2}\left(e^{x} - e^{-x}\right)\right]^4 \\ &= \frac{1}{16}\left(e^{4x} - 4e^{2x} + 6 - 4e^{-2x} + e^{-4x}\right) \end{align*} \]

You should now be able to integrate this term by term...
 

bowlbase

New member
Feb 8, 2012
5
I'm sure I could do it Prove It's way. Whenever it comes to trig problems I just get overwhelmed. Something about identities just throws me for a loop. Even with that information dwsmith, I still can't figure out what to do. Should I replace sinh^4 with u=sinh^2? or should I still replace with u=sinh?
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,184
I'm sure I could do it Prove It's way. Whenever it comes to trig problems I just get overwhelmed. Something about identities just throws me for a loop. Even with that information dwsmith, I still can't figure out what to do. Should I replace sinh^4 with u=sinh^2? or should I still replace with u=sinh?
Neither. dwsmith is recommending that you use the identities he has given you to reduce the powers of the hyperbolic trig functions. If
$$\sinh^{2}(x)=\frac{\cosh(2x)-1}{2},$$
then $\sinh^{4}(x)=?$
 

bowlbase

New member
Feb 8, 2012
5
Neither. dwsmith is recommending that you use the identities he has given you to reduce the powers of the hyperbolic trig functions. If
$$\sinh^{2}(x)=\frac{\cosh(2x)-1}{2},$$
then $\sinh^{4}(x)=?$
So you're saying it would be easier to rewrite it as:

$\sinh^{4}(x)=(\frac{\cosh(2x)-1}{2})^{2}$?
 

sbhatnagar

Active member
Jan 27, 2012
95
Consider \( \displaystyle \boxed{ \displaystyle I_n = \int \sinh^n(x) dx }\)

Note That:

\( \displaystyle \begin{align*} I_n = \int \sinh(x) \cdot \sinh^{n-1}(x) \ dx \end{align*} \)

Use integration by parts:

\( \displaystyle \begin{align*} I_n &=\cosh(x)\cdot\sinh^{n-1}(x)-(n-1)\int \cosh^2(x) \cdot\sin^{n-2}(x) dx \\&=\cosh(x)\cdot\sinh^{n-1}(x)-(n-1)\int \{ 1+\sinh^2(x)\} \sin^{n-2}(x) dx \\ &= \cosh(x)\cdot\sinh^{n-1}(x)-(n-1) \int \sinh^{n-2}(x)+\sinh^n(x) \ dx \\&=\cosh(x)\cdot\sinh^{n-1}(x)-(n-1) \{ I_{n-2}+I_n\}\end{align*} \)

So:

\( \displaystyle I_n= \cosh(x)\cdot\sinh^{n-1}(x)-(n-1) \{ I_{n-2}+I_n\} \)

Solving the above relation for \(I_n \):

\( \displaystyle \boxed{\displaystyle I_n =\frac{\cosh(x)\sinh^{n-1}(x)}{n}-\frac{n-1}{n}I_{n-2}} \)

For \( n=4 \):

\( \displaystyle I_4 =\frac{\cosh(x)\sinh^{3}(x)}{4}-\frac{3}{4}\int \sinh^2(x) dx \)

Apply the identity suggested by dwsmith:

\( \displaystyle I_4 =\frac{\cosh(x)\sinh^{3}(x)}{4}-\frac{3}{4}\int \frac{\cosh(2x)-1}{2} dx \)

Can you finish up now?
 
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DrunkenOldFool

New member
Feb 6, 2012
20
Integration by parts can do wonders!
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,184
So you're saying it would be easier to rewrite it as:

$\sinh^{4}(x)=(\frac{\cosh(2x)-1}{2})^{2}$?
Exactly. Then you can multiply out the RHS of that expression, and use the identity for $\cosh^{2}(x)$, though of course it's really going to be for $\cosh^{2}(2x)$.
 

CaptainBlack

Well-known member
Jan 26, 2012
890
I'm sure I could do it Prove It's way. Whenever it comes to trig problems I just get overwhelmed. Something about identities just throws me for a loop. Even with that information dwsmith, I still can't figure out what to do. Should I replace sinh^4 with u=sinh^2? or should I still replace with u=sinh?
Since the hyperbolic functions are defined as in Prove It's post why are you worried about using hyperbolic trig identities rather than elegant method Prove It gives.

(Note: mathematicians are intrinsically lazy and will always use the method they find easiest or requiring less memorisation. So unless you are asked to use hyperbolic trig for this problem use the simplest method, you might even get more respect that way, and as it is less error prone you might even get more marks)
 

bowlbase

New member
Feb 8, 2012
5
I went ahead and did it Prove It's way. My final answer matched what I was able to get from wolfram so I'm happy with that. However, the difficulty I have with all things trig related really bugs me and I want to strive to be more comfortable with it. I really don't feel compelled to memorize identities as I think memorization is the bane of understanding. That is to say simply memorizing for memorization's sake. Now using identities so often one may incidentally begin remembering them but that is a different thing all together. So this problem will probably slowly go away as time moves on for me since I'm encountering them more often now and will continue to do so through the rest of my physics classes and calculus. But sooner would be better than later.

Anyway, I have not seen this integration by parts method and I think this is along the lines of what wolfram was doing. I'm trying to read up on it now before heading to class. I can't find a reference to it in my calc book though, just seeing wikipedia.