Volume of revolution help?

The region R is bounded by the curve y = 4 +3sec2x, the x-axis and the lines x=-\frac{\pi}{6} and x=\frac{\pi}{6}. Find the volume of the solid formed by rotating R through 2\pi radians about the x-axis.

The region R is bounded by the curve y = 4 +3sec2x, the x-axis and the lines x=-\frac{\pi}{6} and x=\frac{\pi}{6}. Find the volume of the solid formed by rotating R through 2\pi radians about the x-axis.

Are you familiar with this formula: $\pi \int_a^b y^2 \ dx$

Use the pi y^2 formula

The question is: The region R is bounded by the curve y = 4 +3sec2x, the x-axis and the lines x=-pi/6 and x=pi/6. Find the volume of the solid formed by rotating R through 2pi radians about the x-axis.

I get to the stage where I have pi (9/2tan2x + 12ln(sec2x + tan2x) + 16x) with the limits but I cant seem to get the answer in the back when I substitute which is 200.907

You are probably expanding it wrong should get 29

I used the formula (That's how I got to the stage). I'm wondering if I've integrated wrong.

what, 29 for the answer??