# A tricky integral

I'm trying to find the exact value of $$\int_{\frac{1}{\sqrt{3}}}^{\sqrt{3}} \frac{\arctan{(x^2)} }{1+x^2} \, dx$$

Ostensibly, I'd want to use this: $$\frac{d}{dx}\arctan{(x)}=\frac{1}{1+x^2}$$

But either I'm missing something, or this doesn't work out nicely ...

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\begin{align} u & = \frac 1 x \\[8pt] du & = \frac{-dx}{x^2} \\[8pt] \frac{-du}{u^2} & = dx \end{align}

\begin{align} I = & \int_{\frac{1}{\sqrt{3}}}^{\sqrt{3}} \frac{\arctan(x^2)}{1+x^2} \, dx \\[10pt] = & \int_\sqrt{3}^{1/\sqrt{3}} \frac{\arctan\frac{1}{u^2}}{1+\left(\frac{1}{u^2}\right)} \left(\frac{-du}{u^2}\right) \\[10pt] = & \int_{1/\sqrt{3}}^\sqrt{3} \frac{\arctan\left(\frac{1}{u^2}\right)}{u^2+1} \,du \\[10pt] = & \int_{1/\sqrt{3}}^\sqrt{3} \frac{\frac\pi2 - \arctan(u^2)}{u^2+1} \,du \\[10pt] = & \frac \pi 2 \int_{1/\sqrt{3}}^\sqrt{3} \frac{du}{1+u^2} - \int_{1/\sqrt{3}}^\sqrt{3} \frac{\arctan(u^2)}{1+u^2} \,du \\[10pt] = & \frac \pi 2 \int_{1/\sqrt{3}}^\sqrt{3} \frac{du}{1+u^2} - I. \end{align}

So we have $$I = \left(\int\cdots\cdots\cdots\right) - I,$$ whence $$2I = \left(\int\cdots\cdots\cdots\right).$$ That last integral is routine. Remember that $\arctan\sqrt{3} = \pi/3$ and $\arctan(1/\sqrt{3})= \pi/6$.

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bingo! nicely done. +1 –  Bennett Gardiner May 4 at 0:46
@BennettGardiner : Thank you. –  Michael Hardy May 4 at 0:47
@ZubinMukerjee The transformation is strongly suggested by the limits of integration. –  heropup May 4 at 0:52
@ZubinMukerjee : To be a bit more explicit: The bounds of integration are each other's reciprocals, so that's a hint that $u=1/x$ is something to think about. Next, I know that the reciprocal of a random variable with a standard Cauchy distribution also has a standard Cauchy distribution. The Cauchy distribution is the probability distribution $(\text{constant}\cdot dx)/(1+x^2)$. ${}\qquad{}$ –  Michael Hardy May 4 at 3:53
And the fact that $\arctan(1/z) + \arctan(z) =\pi/2$ if $z>0$. –  Bennett Gardiner May 4 at 4:24
Use the substitution $x=\tan\theta$ to obtain: $$I=\int_{\pi/6}^{\pi/3} \arctan(\tan^2\theta)\,d\theta$$ Since $$\int_a^b f(x)\,dx=\int_a^b f(a+b-x)\,dx$$ We get: $$I=\int_{\pi/6}^{\pi/3} \arctan(\cot^2\theta)\,d\theta=\int_{\pi/6}^{\pi/3} \text{arccot}(\tan^2\theta)\,d\theta$$ Add the two to get: $$2I=\int_{\pi/6}^{\pi/3} \frac{\pi}{2}\,d\theta \Rightarrow I=\frac{\pi^2}{24}$$ ....which well agrees with Wolfram Alpha.
Not that busy, I just feel disappointed there. I'll return but not in the near future. In the meantime, I'll have fun here. I,m glad you enjoy my calculus problems. :) Anyway, I wonder how come $\arctan (\cot^2\theta)+\text{arccot}(\tan^2\theta)=\dfrac\pi2$? Are sure about that identity? –  Tunk-Fey May 8 at 2:56
Ah, ok. :) I didn't use the identity you mention, I used $\arctan(\tan^2\theta)+\text{arccot}(\tan^2\theta)=\frac{\pi}{2}$. –  Pranav Arora May 8 at 3:22