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Math Help - Integrating 1/(A+Bx^2)

  1. #1
    Member Jason Bourne's Avatar
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    Integrating 1/(A+Bx^2)

    Dear MHF

    I know that \int{\frac{dx}{A+Bx^2}}=\frac{1}{\sqrt{AB}}\tan^{-1}x\sqrt{\frac{A}{B}}}

    but what is the justification? I guess it's fairly obvious but I can't see it.

    Thanks.
    Jason
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  2. #2
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    Quote Originally Posted by Jason Bourne View Post
    Dear MHF

    I know that \int{\frac{dx}{A+Bx^2}}=\frac{1}{\sqrt{AB}}\tan^{-1}x\sqrt{\frac{A}{B}}}
    but what is the justification?
    Find the derivative of \frac{1}{\sqrt{AB}}\tan^{-1}x\sqrt{\frac{A}{B}}}.
    Do it very carefully step by step.
    You will see how it works.
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  3. #3
    MHF Contributor
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    What you have written is incorrect...

    \displaystyle \int{\frac{dx}{A+Bx^2}}

    Make the substitution \displaystyle x = \sqrt{\frac{A}{B}}\tan{\theta} \implies dx = \sqrt{\frac{A}{B}}\sec^2{\theta}\,d\theta and the integral becomes

    \displaystyle \int{\frac{\sqrt{\frac{A}{B}}\sec^2{\theta}\,d\the  ta}{A + B\left(\sqrt{\frac{A}{B}}\tan{\theta}\right)^2}}

    \displaystyle = \int{\frac{\sqrt{\frac{A}{B}}\sec^2{\theta}\,d\the  ta}{A(1 + \tan^2{\theta})}}

    \displaystyle = \int{\frac{\sqrt{\frac{A}{B}}\sec^2{\theta}\,d\the  ta}{A\sec^2{\theta}}}

    \displaystyle = \int{\frac{\sqrt{\frac{A}{B}}\,d\theta}{\sqrt{A^2}  }}

    \displaystyle = \int{\sqrt{\frac{\frac{A}{B}}{A^2}}\,d\theta}

    \displaystyle = \int{\sqrt{\frac{1}{AB}}\,d\theta}

    \displaystyle = \int{\frac{1}{\sqrt{AB}}\,d\theta}

    \displaystyle = \frac{1}{\sqrt{AB}}\,\theta + C

    \displaystyle = \frac{1}{\sqrt{AB}}\arctan{\left(\sqrt{\frac{B}{A}  }x\right)} + C.
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  4. #4
    Member kjchauhan's Avatar
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    Also you can do as,

    I=\int \frac{dx}{A+Bx^2}

    \therefore I=\frac{1}{A}\int \frac{dx}{1+\left({\frac{\sqrt{B}x}{\sqrt{A}}\righ  t)^2}}

    Now let \frac{\sqrt{B}x}{\sqrt{A}}=t

    \therefore \frac{\sqrt{B}}{\sqrt{A}}dx = dt


    \therefore dx = \frac{\sqrt{A}}{\sqrt{B}}dt

    \therefore I= \frac{\sqrt{A}}{A\sqrt{B}}\int \frac{dt}{1+t^2}

    \therefore I= \frac{1}{\sqrt{AB}} \tan^{-1} t

    \therefore I= \frac{1}{\sqrt{AB}} \tan^{-1} \sqrt{\frac{B}{A}}x
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  5. #5
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    Just to put my oar in, you know, I hope, that the dervative of sin(x) is cos(x) and the derivative of cos(x) is -sin(x).

    From that, and tan(x)= \frac{sin(x)}{cos(x)}, the derivative of tan(x) is \frac{(sin(x))'cos(x)- (sin(x))(cos(x))'}{cos^2(x)}= \frac{cos^2(x)- sin^2(x)}{cos^2(x)}= sec(x).

    Now, if y= arctan(x) then x= tan(y) so that \frac{dx}{dy}= sec^2(y)= sec^2(arctan(x)). But sec^2(\theta)= 1+ tan^2(\theta) so sec^2(arctan(x))= 1+ tan^2(arctan(x))= 1+ x^2

    That is, \frac{dx}{dy}= 1+ x^2 so \frac{dy}{dx}= \frac{1}{1+ x^2}.

    That tells us that \int \frac{dx}{1+ x^2}= arctan(x)+ C
    and the general formula can be derived by substitution as others have shown.
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