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Math Help - [SOLVED] Double check my work?

  1. #1
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    [SOLVED] Double check my work?

    Determine the integral from 0 to 1 of 3/(sqrt(4-x^2)) dx

    Here's what I did:

    3/ sqrt(4(1-(x^2)/4))
    (3/2)(1/(sqrt(1-(x/2)^2))
    3/2 0 to 1 1/(sqrt(1-u^2)) where u=x/2
    3/2 (sin^-1(x))] 0 to 1
    (3/2)*(pi/2)=3pi/4

    Did I do this correctly? If not, could someone please point out my mistakes? Thanks in advance!
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  2. #2
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    Quote Originally Posted by yzobel View Post
    Determine the integral from 0 to 1 of 3/(sqrt(4-x^2)) dx

    Here's what I did:

    3/ sqrt(4(1-(x^2)/4))
    (3/2)(1/(sqrt(1-(x/2)^2))
    3/2 0->1 1/(sqrt(1-u^2)) where u=x/2
    3/2 (sin^-1(x))] 0->1
    (3/2)*(pi/2)=3pi/4

    Did I do this correctly? If not, could someone please point out my mistakes? Thanks in advance!
    \int{\frac{3}{\sqrt{4 - x^2}}\,dx}.

    Use a trigonometric substitution with x = 2\sin{\theta} so that dx = 2\cos{\theta}\,d\theta.

    Then the integral becomes

    \int{\frac{3}{\sqrt{4 - (2\sin{\theta})^2}}\,2\cos{\theta}\,d\theta}

     = \int{\frac{6\cos{\theta}}{\sqrt{4 - 4\sin^2{\theta}}}\,d\theta}

     = \int{\frac{6\cos{\theta}}{\sqrt{4(1 - \sin^2{\theta})}}\,d\theta}

     = \int{\frac{6\cos{\theta}}{2\sqrt{\cos^2{\theta}}}\  ,d\theta}

     = \int{\frac{3\cos{\theta}}{\cos{\theta}}\,d\theta}

     = \int{3\,d\theta}

     = 3\theta + C.


    Now remembering that x = 2\sin{\theta}

    \frac{x}{2} = \sin{\theta}

    \theta = \arcsin{\frac{x}{2}}.


    Therefore, the integral is

    3\theta + C = 3\arcsin{\frac{x}{2}} + C.


    So, putting in the limits...

    \int_0^1{\frac{3}{\sqrt{4 - x^2}}\,dx} = \left[3\arcsin{\frac{x}{2}}\right]_0^1

     = 3\arcsin{\frac{1}{2}} - 3\arcsin{\frac{0}{2}}

     = \frac{3\pi}{6} - 3(0)

     = \frac{\pi}{2}.
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