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Math Help - Integratimg using trig substitution

  1. #1
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    Integratimg using trig substitution

    need some help doing this question..

    \int\frac{x}{\sqrt(4-x^2)}dx given the substitution x=2sint

    using
    4 (1 - \sin^2 x) = 4 cos^2x

    I've managed to rewrite the main integral as

    \int\frac{2sint}{\sqrt(4cos^2t)}dx
    then get stuk
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  2. #2
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    Quote Originally Posted by Kevlar View Post
    need some help doing this question..

    \int\frac{x}{\sqrt(4-x^2)}dx given the substitution x=2sint

    using
    4 (1 - \sin^2 x) = 4 cos^2x

    I've managed to rewrite the main integral as

    \int\frac{2sint}{\sqrt(4cos^2t)}dx
    then get stuk
    Don't forget, if x = 2\sin{t}

    then \frac{dx}{dt} = 2\cos{t}.

    So dx = 2\cos{t}\,dt.


    So your integral becomes

    \int{\frac{x}{\sqrt{4 - x^2}}\,dx} = \int{\frac{2\sin{t}}{\sqrt{4(1 - \sin^2{t})}}\,2\cos{t}\,dt}

     = \int{\frac{4\sin{t}\cos{t}}{2\sqrt{\cos^2{t}}}\,dt  }

     = \int{\frac{2\sin{t}\cos{t}}{\cos{t}}\,dt}

     = \int{2\sin{t}\,dt}

     = -2\cos{t} + C.


    And since x = 2\sin{t}

    x^2 = 4\sin^2{t}

    \frac{x^2}{4} = \sin^2{t}

    \frac{x^2}{4} = 1 - \cos^2{t}

    \cos^2{t} = 1 - \frac{x^2}{4}

    \cos^2{t} = \frac{4 - x^2}{4}

    \cos{t} = \frac{\sqrt{4 - x^2}}{2}.


    So finally, your answer is

    -2\cos{t} + C = \frac{-2\sqrt{4 - x^2}}{2} + C

     =C -\sqrt{4 - x^2}.
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  3. #3
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    Quote Originally Posted by Kevlar View Post
    need some help doing this question..

    \int\frac{x}{\sqrt(4-x^2)}dx given the substitution x=2sint

    using
    4 (1 - \sin^2 x) = 4 cos^2x

    I've managed to rewrite the main integral as

    \int\frac{2sint}{\sqrt(4cos^2t)}dx
    then get stuk
    You have write dx in terms of t.

    dx=2cos(t)dt

    \int\frac{2sint}{\sqrt{4cos^2t}}=\int\frac{4sin(t)  cos(t)}{2cos(t)}dt=2\int sin(t)dt
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