Luckily, the integral is basically set up for you:
[tex]\displaystyle \int_{\theta=0}^{2\pi} \int_{\phi=\frac\pi4}^{\frac\pi2} \int_{\rho=2}^6 \cos(\phi) \, d\rho \, d\phi \, d\theta[/tex]
Since the limits on every variable are constant, and we can factorize [tex]f(\rho,\theta,\phi) = f_1(\rho) f_2(\theta) f_3(\phi)[/tex], we can similarly factorize the integrals. (This is a special case of Fubini's theorem, if I'm not mistaken.)
So the triple integral is equivalent to
[tex]\displaystyle \left(\int_0^{2\pi} d\theta\right) \left(\int_{\frac\pi4}^{\frac\pi2} \cos(\phi) \, d\phi\right) \left(\int_2^6 d\rho\right)[/tex]
and each of these subsequent integrals are easy to compute:
[tex]\displaystyle \int_0^{2\pi} d\theta = \theta \bigg|_0^{2\pi} = 2\pi - 0 = 2\pi[/tex]
[tex]\displaystyle \int_{\frac\pi4}^{\frac\pi2} \cos(\phi) \, d\phi = \sin(\phi) \bigg|_{\frac\pi4}^{\frac\pi2} = \sin\left(\frac\pi2\right) - \sin\left(\frac\pi4\right) = 1 - \frac1{\sqrt2} = \frac{2 - \sqrt2}2[/tex]
[tex]\displaystyle \int_2^6 d\rho = \rho\bigg|_2^6 = 6 - 2 = 4[/tex]
Taken together, the triple integral evaluates to
[tex]\displaystyle \int_{\theta=0}^{2\pi} \int_{\phi=\frac\pi4}^{\frac\pi2} \int_{\rho=2}^6 \cos(\phi) \, d\rho \, d\phi \, d\theta = 2\pi \times \frac{2-\sqrt2}2 \times 4 = \boxed{4\pi(2-\sqrt2)}[/tex]
