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Math Help - Intergration using trig substitution

  1. #1
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    Intergration using trig substitution

    I need to integrate 1/ (1 - x^2)^(1/2) dx

    I know I can use x = sin theta but... I was also taught that I can use x = cos theta

    When I do this I end up with the integral of - sin theta/ sin theta therefore
    the antiderv is negative theta when I know it should be theta!!!! I know I then need to put the answer back in terms of x and add C. Using the sin sub I get arcsin of x + When I use cos sub I get - arccos of x + C!!!!!Could someone please tell me where I am making my mistake? Thanks again for all the time you folks devote! Frostking
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  2. #2
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    Quote Originally Posted by Frostking View Post
    I need to integrate 1/ (1 - x^2)^(1/2) dx

    I know I can use x = sin theta but... I was also taught that I can use x = cos theta

    When I do this I end up with the integral of - sin theta/ sin theta therefore
    the antiderv is negative theta when I know it should be theta!!!! I know I then need to put the answer back in terms of x and add C. Using the sin sub I get arcsin of x + When I use cos sub I get - arccos of x + C!!!!!Could someone please tell me where I am making my mistake? Thanks again for all the time you folks devote! Frostking
    You are not making any mistakes. They are both correct. If we introduce limits into this integral,

    \int^{1}_{-1}\frac{1}{\sqrt{1-x^2}}

    Solution(1), arcsin 1- arcsin(-1) = 180 degrees.

    Solution(2), -arccos 1 + arccos (-1) =180 degrees.

    They both arrive at the same result.
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  3. #3
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    Quote Originally Posted by Frostking View Post
    I need to integrate 1/ (1 - x^2)^(1/2) dx

    I know I can use x = sin theta but... I was also taught that I can use x = cos theta

    When I do this I end up with the integral of - sin theta/ sin theta therefore
    the antiderv is negative theta when I know it should be theta!!!! I know I then need to put the answer back in terms of x and add C. Using the sin sub I get arcsin of x + When I use cos sub I get - arccos of x + C!!!!!Could someone please tell me where I am making my mistake? Thanks again for all the time you folks devote! Frostking
    note the following derivatives ...

    \displaystyle \frac{d}{dx} \arcsin{x} = \frac{1}{\sqrt{1-x^2}}

    \displaystyle \frac{d}{dx} \arccos{x} = -\frac{1}{\sqrt{1-x^2}}

    what's that tell you?
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