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Math Help - Integrals Involving Inverse Trig Functions

  1. #1
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    Integrals Involving Inverse Trig Functions

    Now for integrals involving inverse trig functions.

    A couple problems:

    \int \frac {1}{1 + 4x^2}dx

    For this one, I know that \frac {d}{dx} \arctan {x} = \frac {1}{1-x^2} but I don't know what to do with the 4 in there.

    \int \frac {1}{x^2 - 6x + 13}dx
    This one I have no clue.

    Thanks
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Chicken Enchilada View Post
    Now for integrals involving inverse trig functions.

    A couple problems:

    \int \frac {1}{1 + 4x^2}dx

    For this one, I know that \frac {d}{dx} \arctan {x} = \frac {1}{1-x^2} but I don't know what to do with the 4 in there.

    \int \frac {1}{x^2 - 6x + 13}dx
    This one I have no clue.

    Thanks
    For the first one, note that \int\frac{\,dx}{4x^2+1}=\int\frac{\,dx}{\left(2x\r  ight)^2+1}. Then let u=2x.


    For the second one, note that \int\frac{\,dx}{x^2-6x+13}=\int\frac{\,dx}{(x^2-6x+9)+4}=\int\frac{\,dx}{(x-3)^2+4}=\tfrac{1}{4}\int\frac{\,dx}{\left(\frac{x-3}{2}\right)^2+1}. Then let u=\frac{x-3}{2}.

    Can you take it from here?
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  3. #3
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    Quote Originally Posted by Chris L T521 View Post
    For the first one, note that \int\frac{\,dx}{4x^2+1}=\int\frac{\,dx}{\left(2x\r  ight)^2+1}. Then let u=2x.


    For the second one, note that \int\frac{\,dx}{x^2-6x+13}=\int\frac{\,dx}{(x^2-6x+9)+4}=\int\frac{\,dx}{(x-3)^2+4}=\tfrac{1}{4}\int\frac{\,dx}{\left(\frac{x-3}{2}\right)^2+1}. Then let u=\frac{x-3}{2}.

    Can you take it from here?
    So for the first one the answer is:
    \frac {1}{2} \arctan (2x) + c

    and for the second:
    \frac {1}{2} \arctan (\frac {x-3}{2}) + c

    Sweet! That was actually very simple. You taught me how to solve these problems in a matter of seconds, whereas my prof couldn't do it in 2 hours. It really helps if you can see the problem being solved unambiguously with every little step shown. I think my prof tends to skip a bunch, and everything comes out muddy to me.

    I appreciate the help Chris. Math Help Forum rules!
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