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Math Help - Annoying trigonometric substitution integral

  1. #1
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    Annoying trigonometric substitution integral

    How do you solve this integral: From 0 to a, x^2*sqrt(a^2-x^2)

    The answer is (1/16)a^4pi

    I have no idea where pi comes from, so here's what I do

    Let x=a*sin(theta)

    My work ends up to be (4a^4/32) - (a^4sin4a/32)

    Help!
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  2. #2
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    Re: Annoying trigonometric substitution integral

    When you get your new integral in the theta domain, what is the expression for that?
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  3. #3
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    Re: Annoying trigonometric substitution integral

    Quote Originally Posted by some_nerdy_guy View Post
    How do you solve this integral: From 0 to a, x^2*sqrt(a^2-x^2)

    The answer is (1/16)a^4pi

    I have no idea where pi comes from, so here's what I do

    Let x=a*sin(theta)
    Unfortunately, you error occurs somewhere in here- in exactly the part you did not show. Did you change the limits of integration as you worked? If x= a sin(theta), then when x= 0, a sin(theta)= 0 so theta= 0. When x= a, a= a sin(theta) so sin(theta)= 1, theta= pi/2.

    My work ends up to be (4a^4/32) - (a^4sin4a/32)

    Help!
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  4. #4
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    Re: Annoying trigonometric substitution integral

    Okay, thanks guys. I've figured out the problem, which is, as HallsofIvy stated, I didn't change my limits of integration in which 'a' should have been 'pi/2.' Could anyone possibly explain why if x=a, then theta=pi/2. I don't understand the logic behind that.
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  5. #5
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    Re: Annoying trigonometric substitution integral

    Quote Originally Posted by some_nerdy_guy View Post
    Could anyone possibly explain why if x=a, then theta=pi/2. I don't understand the logic behind that.
    That's because of the substitution, the substitution was x=a\sin(\theta) \Leftrightarrow \theta=\arcsin\left(\frac{x}{a}\right)
    So if x=a that means \theta=\arcsin\left(\frac{a}{a}\right)=\arcsin(1)=  \frac{\pi}{2}
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  6. #6
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    Re: Annoying trigonometric substitution integral

    Ahhh, okay. I really need to brush up on my trig.
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  7. #7
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    Re: Annoying trigonometric substitution integral

    When you do a substitution on a definite integral, you must change over three things: the limits, the integrand, and the differential. For an indefinite integral, of course, while you don't need the limits, you must still transform the integrand and the differential.
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