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Change (9-16cos theta) ^1.5 into another form

  1. Nov 2, 2016 #1
    Mod note: Based on an attachment in a later post in this thread, the actual expression is
    ##(9 - 16\cos^2(\theta))^{3/2}##
    1. The problem statement, all variables and given/known data
    : https://www.physicsforums.com/posts/5610105/ [Broken]

    2. Relevant equations

    3. The attempt at a solution

    my working is (9^1.5) - (16^1.5)[ (cos theta)^1.5 ] = 18-64[ (cos theta)^1.5 ] , is it correct ?
    Last edited by a moderator: May 8, 2017
  2. jcsd
  3. Nov 2, 2016 #2


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    Staff: Mentor

    Try setting ##\theta = \pi/3## in ( 9-16cos theta) ^1.5 and (9^1.5) - (16^1.5)[ (cos theta)^1.5 ]. Do you get the same value?
  4. Nov 2, 2016 #3
    No, how should it be?
  5. Nov 2, 2016 #4


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    Staff: Mentor

    What is ##(a+b)^c##?
  6. Nov 2, 2016 #5
    Dun know , can you help ?
  7. Nov 2, 2016 #6
    here's my actual problem , i wanna integrate $$\int_{0.23\pi}^{0.5\pi} \ (9-16cos\theta)^{1.5}\ d\theta$$
    Mod note: From the attachment, now shown inline, the above should actually be
    ##\int_{0.23\pi}^{0.5\pi} \ (9-16cos^2\theta)^{1.5}\ d\theta##


    Is there any other way to integrate this without expanding the terms ?
    Last edited by a moderator: Nov 2, 2016
  8. Nov 2, 2016 #7

    Ray Vickson

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    Science Advisor
    Homework Helper

    The answer will be a complex number, because ##9 - 16 \cos \theta< 0## over part of the integration region, so when you take its 3/2-power you get an imaginary number. Besides that, your integration is not "elementary": according to Maple, the indefinite integral of ##(a - b \cos\theta)^{3/2}## involves the so-called Elliptic functions.
    Last edited by a moderator: Nov 2, 2016
  9. Nov 2, 2016 #8


    Staff: Mentor

    Thread closed.
    This is what you said:
    The attachment you posted later in this thread shows that the integral is really this: ##\int_{0.23\pi}^{0.5\pi} \ (9-16cos^2\theta)^{1.5}\ d\theta##

    Because of sloppiness on your part, you wasted a fair amount of time of the people trying to help you. Please start a new thread.
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