Proving or disproving an Integral inequality Is the following inequality true ? $$\sqrt[3]{ \int_{0}^{\pi/4} \frac{x}{\cos^2 x \cos^2 (\tan x) \cos^2 \left(\tan (\tan x)\right) \cos^2 \left(\tan \left(\tan \left(\tan x \right)\right)\right)}dx}<\frac{4}{\pi}$$
 A: No.  The trouble comes from the third term in the denominator of the integrand.  The coincidence $\tan(1) \approx \frac{\pi}{2}$ helps explain this:
$\cos(\tan(\tan(\frac{\pi}{4})))^2 = \cos(\tan(1))^2 \approx \cos(\frac{\pi}{2})^2 = 0$
A: Let us write $f^{[k]}$ to denote $\underbrace{f\circ f\circ\cdots\circ f}_{k{\rm times}}$. Then it is easy to see that
$$
\left(\tan^{[4]}(x)\right)'=\frac{1}{\cos^x}\cdot\frac{1}{\cos^2(\tan x)}\cdot
\frac{1}{\cos^2(\tan^{[2]} x)}\cdot\frac{1}{\cos^2(\tan^{[3]} x)}
$$
So the considered integral is $I=\int_0^{\pi/4}x\left(\tan^{[4]}(x)\right)'dx.$
Now, If $h(x)=\tan^{[4]}(x)$ then $h$ has, in the interval $[0,\pi/4]$ , infinitely many zeros $(x_k)_{k\geq 0} $that can be determined explicitly: 
$$
x_k=\arctan^{[2]}\left(\frac{\pi}{2}-\arctan\left(\frac{2}{(2k+1)\pi}\right)\right).
$$
I will consider only $x_0\approx 0.666972$ and $x_1\approx 0.75308$:
$$
I\geq\int_{x_0}^{x_1}xh'(x)dx\geq x_0\int_{x_0}^{x_1}h'(x)dx=+\infty
$$
since $$\lim_{x\to x_0^+}h(x)=-\infty\qquad\hbox{and}\qquad\lim_{x\to x_1^-}h(x)=+\infty$$
Therefore, the considered inequality is not correct.$\qquad\square$
