Thread: can't integrate powers of hyperbolics

1. can't integrate powers of hyperbolics

Hey guys, I'm having touble inegrating:

(cosh[3t]^5)(sinh[3t]^4)

I've tried a number of things, integration by parts, using the identities and double angle formulae, substitution and a bunch of combinations of the above but cannot get the result given by The Integrator--Integrals from Mathematica which gives a hideous looking beast of an answer.

and i also can't do:

(u+3)/([u^2]+4u+8)

I believe I need to complete the square and then use a trigonometric substitution but can't figure it out, or maybe I'm getting lazy. Insight please?

Would appreciate any help.. thanks guys.

Aleks

2. For the second one, completing the square is a good approach, but you don't have to use trig sub.

Upon completing the square, we get:

$\displaystyle \int\frac{x+3}{(x+2)^{2}+6}dx$

Let $\displaystyle u=x+2, \;\ du=dx, \;\ u+1=x+3$

$\displaystyle \int\frac{u+1}{u^{2}+4}du=\int\frac{u}{u^{2}+4}du+ \int\frac{1}{u^{2}+4}du$

I believe you have an ln and an arctan in your future.

3. That's great, thanks! But I prevailed with that probllem earlier (with the trig sub). I'm still stuck on the hyperbolics though, that one seems impossible would appreciate any more input. Thanks.

Aleks

4. There's probably easier ways, but let's use the e identities.

We can expand $\displaystyle cosh^{5}(3t)$ and $\displaystyle sinh^{4}(3t)$ to their equaivalent e idneitites and then integrate term by term.

$\displaystyle cosh^{5}(3t)sinh^{4}(3t)=$

$\displaystyle \frac{1}{512}e^{27t}+\frac{1}{512}e^{21t}-\frac{1}{128}e^{15t}-\frac{1}{128}e^{9t}$$\displaystyle +\frac{3}{256}e^{3t}+\frac{3}{256}e^{-3t}-\frac{1}{128}e^{-9t}-\frac{1}{128}e^{-15t}+\frac{1}{512}e^{-21t}+\frac{1}{512}e^{-27t}$

Now, it's long, but easy to integrate term by term.

5. Hello, Aleks!

$\displaystyle \int \cosh^5(3t)\,\sinh^4(3t)\,dt$
You're expected to know: .$\displaystyle \cosh^2x - \sinh^2x \:=\:1\quad\Rightarrow\quad \cosh^2x \:=\:\sinh^2x + 1$

Let $\displaystyle \theta = 3t$

We have: .$\displaystyle \cosh^4\!\theta\cdot\sinh^4\!\theta\cdot\cosh\thet a \;=\;(\cosh^2\!\theta)^2\cdot\sinh^4\!\theta\cdot\ cosh\theta \;= \;\left(\sinh^2\!\theta + 1\right)^2\cdot\sinh^4\!\theta\cdot\cosh\theta$

. . $\displaystyle = \;\left(\sinh^4\!\theta + 2\sinh^2\!\theta + 1\right)\cdot\sin^4\!\theta\cdot\cosh\theta \;=\;\left(\sinh^8\!\theta + 2\sinh^6\!\theta + \sinh^4\!\theta\right)\cdot\cosh\theta$

Let: $\displaystyle u \,= \,\sinh\theta\quad\Rightarrow\quad du \,=\,\cosh\theta\,d\theta$

And we have: .$\displaystyle \int\left(u^8 + 2u^6 + u^4\right)\,du$

Can you finish it now?

6. Originally Posted by aleksr
Hey guys, I'm having touble inegrating:

(cosh[3t]^5)(sinh[3t]^4)
$\displaystyle \int \cosh^5 3t \sinh^5 3t dt$

$\displaystyle \cosh^5 3t \sinh^4 3t = \cosh^4 3t \sinh^4 t \cosh 3t$
$\displaystyle (1 + \sinh^2 3t)\sinh^4 t \cosh 3t$

Use the substitution $\displaystyle x=\sinh 3t \implies x'= \cosh 3t$