# Integrate $\int_{0}^{\frac{\pi}{4}}\tan^{-1}\left(\frac{\sqrt{2}\cos3 \phi}{\left(2\cos 2 \phi+ 3\right)\sqrt{\cos 2 \phi}}\right)d\phi$

Evaluate the integral:

$$\displaystyle \int_{0}^{\frac{\pi}{4}}\tan^{-1}\left(\frac{\sqrt{2}\cos3 \phi}{\left(2\cos 2 \phi+ 3\right)\sqrt{\cos 2 \phi}}\right)d\phi$$

I have no clue on how to attack it.
The only thing I noticed is that there exists a symmetry around $\pi/8$, meaning that from $\pi/8$ to $\pi/4$ is the negative of zero to $\pi/4$. But, there exists a root of the integrand at $\pi/6$ and the limit of the integrand at $\pi/4$ is $-\infty$.

Conjecture: The integral is $0$ for the reason of symmetry I mentioned above.

However I cannot prove that. I would appreciate your help.

#### Solutions Collecting From Web of "Integrate $\int_{0}^{\frac{\pi}{4}}\tan^{-1}\left(\frac{\sqrt{2}\cos3 \phi}{\left(2\cos 2 \phi+ 3\right)\sqrt{\cos 2 \phi}}\right)d\phi$"

By replacing $\phi$ with $\arctan(t)$, then using integration by parts, we have:

$$I = \int_{0}^{1}\frac{1}{1+t^2}\,\arctan\left(\frac{\sqrt{2}(1-3t^2)}{(5+t^2)\sqrt{1-t^2}}\right)\,dt =\frac{\pi^2}{8}-\int_{0}^{1}\frac{3\sqrt{2}\, t \arctan(t)}{(3-t^2)\sqrt{1-t^2}}\,dt.$$
Now comes the magic. Since:
$$\int \frac{3\sqrt{2}\,t}{(3-t^2)\sqrt{1-t^2}}\,dt = -3\arctan\sqrt{\frac{1-t^2}{2}}\tag{1}$$
integrating by parts once again we get:

$$I = \frac{\pi^2}{8}-3\int_{0}^{1}\frac{1}{1+t^2}\arctan\sqrt{\frac{1-t^2}{2}}\,dt \tag{2}$$
hence we just need to prove that:
$$\int_{0}^{1}\frac{dt}{1+t^2}\,\arctan\sqrt{\frac{1-t^2}{2}}=\int_{0}^{\frac{1}{\sqrt{2}}}\frac{\arctan\sqrt{1-2t^2}}{1+t^2}\,dt=\color{red}{\frac{\pi^2}{24}}\tag{3}$$
and this is not difficult since both
$$\int_{0}^{1}\frac{dt}{1+t^2}(1-t^2)^{\frac{2m+1}{2}},\qquad \int_{0}^{\frac{1}{\sqrt{2}}}\frac{(1-2t^2)^{\frac{2m+1}{2}}}{1+t^2}\,dt$$
can be computed through the residue theorem or other techniques. For instance:
$$\int_{0}^{1}\frac{(1-t)^{\frac{2m+1}{2}}}{t^{\frac{1}{2}}(1+t)}\,dt = \sum_{n\geq 0}(-1)^n \int_{0}^{1}(1-t)^{\frac{2m+1}{2}} t^{n-\frac{1}{2}}\,dt=\sum_{n\geq 0}(-1)^n\frac{\Gamma\left(m+\frac{3}{2}\right)\Gamma\left(n+\frac{1}{2}\right)}{\Gamma(m+n+2)}$$
or just:
$$\int_{0}^{1}\frac{\sqrt{\frac{1-t^2}{2}}}{(1+t^2)\left(1+\frac{1-t^2}{2}u^2\right)}\,dt = \frac{\pi}{2(1+u^2)}\left(1-\frac{1}{\sqrt{2+u^2}}\right)\tag{4}$$
from which:
$$\int_{0}^{1}\frac{dt}{1+t^2}\,\arctan\sqrt{\frac{1-t^2}{2}}=\frac{\pi}{2}\int_{0}^{1}\frac{du}{1+u^2}\left(1-\frac{1}{\sqrt{2+u^2}}\right) =\color{red}{\frac{\pi^2}{24}}$$
as wanted, since:
$$\int \frac{du}{(1+u^2)\sqrt{2+u^2}}=\arctan\frac{u}{\sqrt{2+u^2}}.$$