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Math Help - Inverse Trigonometry.

  1. #1
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    Exclamation Inverse Trigonometry.

    A curve with gradient function (1-x)/(1+x^2) goes through the origin. Show that the curve cuts the X-axis when arctan(x) = 0.5ln(1+x^2)

    What exactly am I meant to do for these types of questions?
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  2. #2
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by Sunyata View Post
    A curve with gradient function (1-x)/(1+x^2) goes through the origin. Show that the curve cuts the X-axis when arctan(x) = 0.5ln(1+x^2)

    What exactly am I meant to do for these types of questions?
    To find the original function you need to integrate

    \int\frac{1-x}{1+x^2}\,dx=\int\frac{1}{1+x^2}\,dx-\int\frac{x}{1+x^2}\,dx

    The first integral is \arctan(x) (you should find this on any table of integrals) and the second integral can be solved using a simple u-sub ( u=1+x^2), so you end up with f(x)=\arctan(x)-\frac{1}{2}\ln(1+x^2). Then you set f(x)=0 to get the answer.

    It seems a little strange that the question would ignore the presence of the integration constant though.
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