Integrals and comparison Let $I_{1} = \int_0^\frac{\pi}{4}  (\tan x)^{\frac{1}{3}}dx$  and 
$I_{2} = \int_0^\frac{\pi}{4} (\tan x)^{\sqrt 2}dx $ then....
Prove that $I_{1}\lt \frac{3}{4}$ , $I_{2}\lt \frac{\pi}{4}$ ....
and $I_{2}\lt \log(\sqrt 2) \lt I_{1}$
So basically i assumed in $I_{1}$ that $\tan x= t^\frac{3}{2}\\$ ending up as $\frac{3}{2} \int_0^1 \frac{t}{t^3 +1}dt$ but could not think ahead..
 A: There is an interesting Lemma for dealing with such kind of integrals.

Lemma. If $I=(a,b)$ and $f(x),\omega(x)$ are non-negative, bounded and continuous on $I$, the (moment) function
  $$ M(s) = \int_{I} f(x)^s \omega(x)\,dx $$
  is non-negative and log-convex on $\mathbb{R}^+$ due to the Cauchy-Schwarz inequality.
  In particular it is a convex function on $\mathbb{R}^+$.

In our case we have $I=\left(0,\frac{\pi}{4}\right)$,  $f(x)=\tan(x)$ and $\omega(x)=1$.
It is straightforward to check that $M(0)=\frac{\pi}{4}$, $M(1)=\log\sqrt{2}$, $M(2)=1-\frac{\pi}{4}$.
Moreover $M(s)$ is decreasing on $\mathbb{R}^+$, since on $I$ we have $0<f(x)<1$.
These observations immediately settle $I_2<\frac{\pi}{4}$ and $I_2<M(1)<I_1$.
And by convexity
$$ I_1=M\left(\frac{1}{3}\right)\leq \frac{2}{3}M(0)+\frac{1}{3}M(1)=\frac{\pi+\log(2)}{6}<\frac{3}{4}.$$

An explicit computation of $I_1$ can be carried out in the following way:
$$ I_1 = \int_{0}^{1}\frac{t^{1/3}}{1+t^2}\,dt = 3\int_{0}^{1}\frac{s^3}{1+s^6}\,ds $$
and by partial fraction decomposition the last integral equals $\frac{\pi}{2\sqrt{3}}-\frac{\log 2}{2}\approx 0.56$.
We may also notice that by the Cauchy-Schwarz inequality
$$ I_1 \leq \sqrt{\int_{0}^{1}t^{-1/3}\,dt \int_{0}^{1}\frac{t\,dt}{(1+t^2)^2}} = \sqrt{\frac{3}{8}}.$$
