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Thread: Simple Integral

  1. #1
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    Simple Integral

    Integration of 1/(1 + x^2)

    when I use x = tan x, I get the answer as tan^-1(x)

    when I use x = cot x, I get the answer as -cot^-1(x)

    Does this mean tan^-1(x) = -cot^-1(x)? How is this possible?
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  2. #2
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    Re: Simple Integral

    Quote Originally Posted by joshuaa View Post
    Integration of 1/(1 + x^2)

    when I use x = tan x, I get the answer as tan^-1(x)

    when I use x = cot x, I get the answer as -cot^-1(x)

    Does this mean tan^-1(x) = -cot^-1(x)? How is this possible?
    Don't you know that $D_x(\arctan(x))=\dfrac{1}{1+x^2}~\&~D_x(\cot^{-1}(x)=\dfrac{-1}{1+x^2}~?$
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  3. #3
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    Re: Simple Integral

    Quote Originally Posted by joshuaa View Post
    Integration of 1/(1 + x^2)

    when I use x = tan x, I get the answer as tan^-1(x)

    when I use x = cot x, I get the answer as -cot^-1(x)

    Does this mean tan^-1(x) = -cot^-1(x)? How is this possible?
    $\arctan(x)$ and $-\text{arccot}(x)$ differ by a constant ...

    $\arctan(x) = \dfrac{\pi}{2}-\text{arccot}(x)$
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  4. #4
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    Re: Simple Integral

    Quote Originally Posted by joshuaa View Post
    Integration of 1/(1 + x^2)

    when I use x = tan x, I get the answer as tan^-1(x)

    when I use x = cot x, I get the answer as -cot^-1(x)
    You mean, I hope, "u= tan(x)" and "u= cot(x)". And you should have got tan^{-1}(x)+ C and cot^{-1}(x)+ C. As others have said, yes, those are the same with different constants.


    Does this mean tan^-1(x) = -cot^-1(x)? How is this possible?
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  5. #5
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    Re: Simple Integral

    Quote Originally Posted by Plato View Post
    Don't you know that $D_x(\arctan(x))=\dfrac{1}{1+x^2}~\&~D_x(\cot^{-1}(x)=\dfrac{-1}{1+x^2}~?$
    Just realized now.
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    Re: Simple Integral

    thanks guys
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