Integrating 1/(A+Bx^2)

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February 26th 2011, 03:41 AM
Jason Bourne

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
February 26th 2011, 04:10 AM
Plato

Quote:

Originally Posted by Jason Bourne

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.
February 26th 2011, 05:13 AM
Prove It

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

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

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

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

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

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

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

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

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

$\displaystyle = \frac{1}{\sqrt{AB}}\arctan{\left(\sqrt{\frac{B}{A} }x\right)} + C$ .
February 26th 2011, 04:08 PM
kjchauhan

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 $\frac{\sqrt{B}x}{\sqrt{A}}=t$

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

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

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

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

$\therefore I= \frac{1}{\sqrt{AB}} \tan^{-1} \sqrt{\frac{B}{A}}x$
February 27th 2011, 04:43 AM
HallsofIvy

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.