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Calculus II - Trigonometric Integrals HARD

  1. Jul 30, 2011 #1
    1. The problem statement, all variables and given/known data

    Evaluate

    integral sin^(-3/2)(x)cos^3(x) dx

    2. Relevant equations

    tan(x)=sin(x)/cos(x)
    sin^2(x)+cos^2(x)=1
    sin^2(x)=(1-cos(x))/2
    cos^2(x)=(1+cos(2x))/2
    integral cos(x)dx = sin(x) + c
    integral sin(x)dx = -cos(x) + c
    d/dx sin(x) = cos(x)
    d/dx cos(x) = -sin(x)
    a^m/a^n=a^(m-n)
    a^m*a^n=a^(m+n)
    sin(x)=(e^(ix)-e^(-ix))/(2i)
    cos(x)=(e^(ix)+e^(-ix))/2
    tan(x)=sin(x)/cos(x)
    cis(x)=cos(x)+i*sin(x)=e^(ix)
    cis(-x)=cos(x)-i*sin(x)=e^(-ix)
    e^(i*pi)+1=0

    3. The attempt at a solution
    I'm at a lost as to how even to being...
    I tried using sin^2(x)+cos^2(x)=1
    I tried using some of the double angle formulas
    every single time I get to the point were I don't know how to proceed

    MATLAB answer:
    >> int(cos(x)^3/sin(x)^(3/2))
    Warning: Explicit integral could not be found.

    ans =

    int(cos(x)^3/sin(x)^(3/2), x)

    Wolfram Alpha answer:
    -40/9 F(1/4 (pi-2 x)|2)-2/9 sqrt(sin(x)) (2 cos(x)+3 x (sin(x)+3 csc(x)))+constant

    My answer:
    Don't even know how to proceed after a couple of steps

    Back of the book:
    Only odd answers are given... this is problem #16
     
    Last edited: Jul 30, 2011
  2. jcsd
  3. Jul 30, 2011 #2

    Ray Vickson

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    Maple gets a simple answer:
    > f:=cos(x)^3/sin(x)^(3/2);
    > J2:=int(f,x);
    > J2 := -2/3*(sin(x)^2+3)/sin(x)^(1/2)

    We can check to see if dJ2/dx = f:
    > simplify(diff(J2,x));

    > cos(x)^3/sin(x)^(3/2) <---- correct!

    RGV
     
  4. Jul 30, 2011 #3
    so how do i go about proceeding in this problem?
    as far as doing it by hand
     
  5. Jul 30, 2011 #4

    Ray Vickson

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    Try sin(x)=y.

    RGV
     
  6. Jul 30, 2011 #5

    I like Serena

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  7. Jul 30, 2011 #6

    eumyang

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    What is the original question? The OP wrote it two different ways. Is the integrand this?
    [tex]\sin^{-3/2} x \text{ }\cos^3 x[/tex]
    Or this?
    [tex]\frac{\cos^3 x}{\sin^{-3/2} x}[/tex]
     
  8. Jul 30, 2011 #7
    Greenprint, there's a simple first step for these sorts of problems.. first:

    [tex] \int sin^{-3/2}(x)cos^{3}(x)dx = \int sin^{-3/2}(x)(1-sin^{2}(x))cos(x)dx [/tex]

    Can you see a substitution you can make now that might make things easier?

    edit - eumyang, look at how the fraction version is written. the -3/2 is turned into a 3/2 by putting it in the denominator.
     
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