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Math Help - trigonometry substitution

  1. #1
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    trigonometry substitution

    \int \frac {1}{x^2sqrt(1-x^2)} dx

    x = sin t
    dx = cos t dt

    = \int \frac{cos t}{sin^2tsqrt(1-sin^2t)}

    = \int \frac {cos t}{sin^2tsqrt(cos^2t)}

    = \int \frac {1}{sin^2t}

    That's as far as I got ... how can I proceed?

    PS; I apologize for the untidiness of the square roots. How do I make the radical appear?
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  2. #2
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    Quote Originally Posted by Archduke01 View Post
    \int \frac {1}{x^2sqrt(1-x^2)} dx

    x = sin t
    dx = cos t dt

    = \int \frac{cos t}{sin^2tsqrt(1-sin^2t)}

    = \int \frac {cos t}{sin^2tsqrt(cos^2t)}

    = \int \frac {1}{sin^2t}

    That's as far as I got ... how can I proceed?

    PS; I apologize for the untidiness of the square roots. How do I make the radical appear?
    \frac {1}{sin^2t}=csc^2t
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  3. #3
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    OK, so I finally end up with - cot t + C

    I'm confused with making the triangle though. I have my x = sin t, but using that, all my sides will basically be 1. Is that right?
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  4. #4
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    Quote Originally Posted by Archduke01 View Post
    I'm confused with making the triangle though. I have my x = sin t, but using that, all my sides will basically be 1. Is that right?
    t is an acute angle of the triangle. The side opposite t is x, the hypotenuse is 1, and the side adjacent to t is \sqrt{1-x^2}
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  5. #5
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    Are you evaluating this over an interval or is it an indefinite integral?

    The method you use is a bit tricky if it is not over a small interval (i.e., one where cos x is always positive or negative) because:

    \frac{\cos x}{\sqrt{\cos^2 x}} = \frac{\cos x}{|\cos x|} = \begin{cases} 1, & \cos x > 0 \\ -1, & \cos x < 0\end{cases}

    P.S. Mouse over the math symbols to see the code used to generate them.

    You could also do this problem using u=\sqrt{1-x^2}
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