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I would like to show that for $A(x) = \int_{0}^{x}\frac{1}{1+t^2}dt$, we have $A\left(\frac{2x}{1-x^2}\right)=2A(x)$, for all $|x|<1$.

My idea is to start with either $2\int_0^x\frac{1}{1+t^2}dt$ or $\int_0^{2x/(1-x^2)}\frac{1}{1+t^2}dt$, and try to transform one into the other by change of variables. (It would make more sense for the moment if we did not do any trigonometric substitutions, since we are defining the trig functions via this integral.)

One of the several things I've tried is to use $A(x)+A(1/x)=\pi/2$, and write $\int_0^{2x/(1-x^2)}\frac{1}{1+t^2}dt=\pi/2 - \int_0^{(1-x^2)/2x}\frac{1}{1+t^2}dt=\pi/2 - \int_0^{1/2x}\frac{1}{1+t^2}dt+\int_0^{x/2}\frac{1}{1+t^2}dt$, but this doesn't seem to be getting me anywhere.

Any ideas?

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Does it have to be a change of variables, or are other techniques (not using trigonometry) okay? –  Daniel Fischer Jul 14 '13 at 21:04
    
Sorry for the vagueness; any technique is fine. –  Eric Auld Jul 14 '13 at 21:06
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Ah, well, it's already posted. –  Daniel Fischer Jul 14 '13 at 21:07
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The quiz at 7:30 AM on Monday will ask you to prove that $\displaystyle3\int_0^x\frac{dt}{1+t^2} = \int_0^{\large \frac{3x-x^3}{1-3x^2}} \frac{dt}{1+t^2}$. ${}\qquad{}$ –  Michael Hardy Jul 14 '13 at 23:05

1 Answer 1

up vote 8 down vote accepted

Hint: Let $f(x)=A\left(\frac{2x}{1-x^2}\right)$ and let $g(x)=2A(x)$, both defined by integrals precisely as in the OP.

Use the Fundamental Theorem of Calculus to show that these functions have the same derivative. So they differ by a constant. Then all you will need to do is to show that they agree at say $x=0$, so the constant is $0$.

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This method doesn't answer the question. –  Sami Ben Romdhane Jul 14 '13 at 21:09
    
@SamiBenRomdhane: I believe that it does. After some calculation, we find that $f'(x)=\frac{2}{1+x^2}$. –  André Nicolas Jul 14 '13 at 21:12
    
In fact the OP ask a method by using the integral definition of $\arctan$ and your method is based on the derivation. –  Sami Ben Romdhane Jul 14 '13 at 21:15
    
@SamiBenRomdhane But from the integral definition one gets that $(\arctan x)'=(1+x^2)^{-1}$. –  Pedro Tamaroff Jul 14 '13 at 21:16
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@EricAuld: You are welcome. So the calculation went smoothly! The reason for the restriction is that $\frac{2x}{1-x^2}$ blows up at $\pm 1$, so beyond that we cannot use the theorem that if derivatives agree then functions differ by a constant. –  André Nicolas Jul 14 '13 at 21:25

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