Simplifying $\sqrt{1+\cos^2(x)}$ for various values of x I'm being asked to find the arc length of $y=\sin( x)$ for $[0, \frac{\pi}{2}]$ using $M_8$.
I've determined that $y\prime^2=\cos^2x$.  So, using the formula for arc length, I get $\sqrt{1+\cos^2x}$ as my function.
Now, they want me to evaluate this using $M_8$, so I end up with $8$ midpoints from $\frac{\pi}{32}$ to $\frac{15\pi}{32}$ moving up in increments of  $\frac{\pi}{16}$.
Also, I know that $\Delta{x} = \frac{\pi}{16}$.  So, I end up with this estimated integral:
$$\frac{\pi}{16}\left(\sqrt{1+\cos^2(\tfrac{\pi}{32})} + \sqrt{1+\cos^2(\tfrac{3\pi}{32})} + ... \sqrt{1+\cos^2(\tfrac{15\pi}{32})}\right) = 1.91009889$$
Please feel free to correct any errors here.  
My question, however, has to do with simplifying this answer using trig.  I would assume that there's a fancy way to do this, rather than just handing in the answer $1.91009889$ (though the prof will accept this).  
I'd like to sharpen my trig skills, so I figured this was a great chance.

Offhand, not much comes to mind with regard to $\sqrt{1+\cos^2x}$, but $\cos\frac{k\pi}{2^n}$ can be evaluated exactly using the half-angle identity $\cos\frac{x}{2}=\pm\sqrt{\frac{1+\cos x}{2}}$ (choose + or – based on the value of $\frac{x}{2}$) repeatedly.
