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Math Help - can't integrate powers of hyperbolics

  1. #1
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    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
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  2. #2
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    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:

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

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

    \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.
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  3. #3
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    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
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  4. #4
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    There's probably easier ways, but let's use the e identities.

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

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

    \frac{1}{512}e^{27t}+\frac{1}{512}e^{21t}-\frac{1}{128}e^{15t}-\frac{1}{128}e^{9t} +\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.
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  5. #5
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    Hello, Aleks!

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


    Let \theta = 3t

    We have: . \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

    . . = \;\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: u \,= \,\sinh\theta\quad\Rightarrow\quad du \,=\,\cosh\theta\,d\theta

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


    Can you finish it now?

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

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

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

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