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Thread: [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!
    $\displaystyle \int{\frac{3}{\sqrt{4 - x^2}}\,dx}$.

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

    Then the integral becomes

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

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

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

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

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

    $\displaystyle = \int{3\,d\theta}$

    $\displaystyle = 3\theta + C$.


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

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

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


    Therefore, the integral is

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


    So, putting in the limits...

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

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

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

    $\displaystyle = \frac{\pi}{2}$.
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