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Integration by Parts with a Touch of Trig

  1. Jan 30, 2007 #1
    Hey y'all. I'm new to the forum, and have a problem that I've been working on all night long. I'm having issues previewing the Latex, so bear with me. I'll post the work I've done so far if the problem code shows up. Thanks.

    [tex]\int{\sin^{\frac{3}{2}}2\theta\cos^{3}2\theta} d\theta[/tex]

    Now, the fraction power in the sin is really throwing me off. I've tried rewriting the problem like this:

    [tex]\int{\sqrt{\sin^{3}2\theta}\cos^{3}2\theta} d\theta[/tex]

    Singled out the sine in the square root:

    [tex]\int{\sqrt{\sin2\theta\sin^{2}2\theta}\cos^{3}2\theta} d\theta[/tex]

    Square rooted the squared sine function:

    [tex]\int{\sqrt{\sin2\theta}\sin2\theta\cos^{3}2\theta} d\theta[/tex]

    Skipping some steps (summarized): I split the cosine into two parts ([tex]\cos^{2}2\theta \cos2\theta[/tex]) and then used the trig identity to change [tex]\cos^{2}2\theta[/tex] to [tex](1-\sin^{2}2\theta)[/tex]. Put everything together and got this:

    [tex]\int{\sqrt{\sin2\theta}\sin2\theta\cos2\theta-sin^{3}2\theta\cos2\theta d\theta}[/tex]

    And now I'm completely lost. I'm not sure if I'm moving in the right direction or just making it a more difficult problem then it really is.

    Now that I think about it, I'm not sure if I use integration by parts here.
    Last edited: Jan 30, 2007
  2. jcsd
  3. Jan 30, 2007 #2
    The easiest way I can think about doing this integral requires a bit of a trick. Have you ever seen the proof as to why

    [tex]\int e^{x^2} dx =\sqrt{2\pi}[/tex] ?

    We will use the same trick here. As a matter of fact, you yourself have said that you don't like the fractional power, so let's get rid of it.

    Let [tex]f(x)=\int{\sin^{\frac{3}{2}}2\theta\cos^{3}2\theta} d\theta[/tex]

    Then [tex] [f(x)]^2=\int{\sin^3 2\theta \cos^6 2\theta} d\theta[/tex]

    We'll save one sine, and then use the identity

    [tex]sin^2 2\theta = 1-cos^2 2\theta[/tex]

    this gives

    [tex] [f(x)]^2=\int{\sin 2\theta \sin^2 2\theta \cos^6 2\theta} d\theta[/tex]

    [tex] [f(x)]^2=\int{\sin 2\theta (1-cos^2 2\theta) \cos^6 2\theta} d\theta[/tex]

    [tex] [f(x)]^2=\int{\sin 2\theta ( \cos^6 2 \theta-cos^8 2\theta) d\theta[/tex]

    now make the substitution [tex] u = cos 2\theta [/tex] and everything follows easily from there. Once you have your solution, simply substitute back to trigonometric functions, and take the square root to return to [tex]f(x)[/tex].
  4. Jan 30, 2007 #3
    Thanks for the help. I ran into another brick wall though. I worked the problem down to the part where I get this:

    [tex] [f(x)]^2=\int{\sin 2\theta ( \cos^6 2 \theta-cos^8 2\theta) d\theta[/tex]

    And substitute:

    u: [tex] u = \cos 2\theta [/tex]

    du: [tex] du = -2\sin 2\theta [/tex]

    du: [tex] -\frac{1}{2}du = \sin 2\theta[/tex]

    So I'm left with the following integral:

    [tex] -\frac{1}{2} \int { u^6 - u^8 } du[/tex]

    Which I integrate and get the following:

    [tex] -\frac{1}{2} ( \frac{1}{7} u^7 - \frac{1}{9} u^9 ) \\
    =-\frac{1}{14} u^7 + \frac{1}{18} u^9[/tex]

    I then squareroot the entire function, because we squared is earlier. And I'm left with:

    [tex]-\frac{\sqrt{14}}{14} cos^\frac{7}{2}2\theta + \frac{\sqrt{2}}{6} cos^\frac{9}{2}2\theta + C[/tex]

    This looks really ugly, and when I plug it into a graph, it doesn't match up. I even tried plugging it in and simplifying it in my CAS and still no luck. I'm not asking for the answer, I just want to know if I goofed up somewhere.
    Last edited: Jan 30, 2007
  5. Jan 30, 2007 #4
    It looks like I goofed up the Latex too... I can't seem to fix it.
  6. Jan 30, 2007 #5


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    Staff Emeritus
    Science Advisor

    Is this what you meant? (You missed a bracket after \sqrt{2})

    [tex]-\frac{\sqrt{14}}{14} cos^\frac{7}{2}2\theta + \frac{\sqrt{2}}{6} cos^\frac{9}{2}2\theta + C[/tex]
  7. Jan 30, 2007 #6
    Yea, that's it. My latex image won't refresh.
  8. Jan 30, 2007 #7
    Remember, [tex]\sqrt{x+y} \neq \sqrt{x}+\sqrt{y} [/tex], in general
  9. Jan 31, 2007 #8


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    Staff Emeritus
    Science Advisor

    You meant, of course,
    [tex]\int_0^\infty e^{-x^2}dx= \sqrt{2\pi}[/tex]

    Are you sure that's true? That is NOT the "trick" that is used in the [itex]e^{-x^2}[/itex] integration you refered to.

    If [itex]f(x)= \int x dx= \frac{1}{2}x^2[/itex], then [itex][f(x)]^2= \frac{1}{4}x^4[/itex] which is NOT [itex]\int x^2 dx[/itex]!
  10. Feb 1, 2007 #9
    Thank you for the correction to the [tex]e^(-x^2)[/tex]; however it is indeed the trick used to solve that integral, though I admit that I did not apply it properly in this case.

    The trick is indeed to square the function, giving you a product of two integrals of independent variables x and y, then to make a change of co-ordinates to a polar representation so that the Jacobian gives us an elementary anti-derivative.

    For your example

    [itex]f(x)= \int x dx= \frac{1}{2}x^2[/tex]
    [tex][f(x)]^2 = \displaystyle \int \int xy \,dx dy [/tex]
    [tex]=\int \frac{1}{2}x^2 y dy [/tex]
    [tex]\Rightarrow f(x) = \frac{1}{2} x*y [/tex]

    Which, when [tex]x=y[/tex] is indeed the correct solution.
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