How to Solve the Arctan Integral Equal to π^2/20?

[Tony1]

Show that,

$$\int_{0}^{\pi/6}\tan^{-1}\sqrt{2-\tan^2(x)}\mathrm dx={\pi^2\over 20}$$

 

[jvicens]

Hello everyone,

I came across this interesting integral while working on a research project and I'm having trouble solving it. Can someone please help me out?

To solve this integral, we can use the substitution $u = \tan(x)$. This will change the limits of integration to $u = 0$ and $u = \sqrt{2}$, and the integral becomes:

$$\int_{0}^{\sqrt{2}}\tan^{-1}\sqrt{2-u^2}\mathrm du$$

Next, we can use the identity $\tan^{-1}(x) = \frac{1}{2}\tan^{-1}(\frac{2x}{1-x^2})$ to rewrite the integrand as:

$$\frac{1}{2}\int_{0}^{\sqrt{2}}\tan^{-1}\left(\frac{2\sqrt{2-u^2}}{1-(2-u^2)}\right)\mathrm du$$

Simplifying, we get:

$$\frac{1}{2}\int_{0}^{\sqrt{2}}\tan^{-1}\left(\frac{2\sqrt{2-u^2}}{u^2-1}\right)\mathrm du$$

Next, we use the substitution $v = \frac{2\sqrt{2-u^2}}{u^2-1}$, which gives us $dv = \frac{2}{(u^2-1)^2}\mathrm du$ and changes the limits of integration to $v = \frac{2\sqrt{2}}{3}$ and $v = \frac{2\sqrt{2}}{2}$.

The integral now becomes:

$$\frac{1}{4}\int_{\frac{2\sqrt{2}}{3}}^{\frac{2\sqrt{2}}{2}}\tan^{-1}(v)\mathrm dv$$

Using the formula for the integral of $\tan^{-1}(x)$, we get:

$$\frac{1}{4}\left[\left(\frac{1}{2}v\tan^{-1}(v)-\frac{1}{4}\ln(1+v^2)\right)\right]_{\frac{2\sqrt{2}}{3}}^{\frac{2\sqrt{2}}{2}}$$

Simplifying and evaluating at the limits of integration