Sec-cubed

$\int_{0}^{\frac{\pi}{3}}\sec^{3}x\textrm{d}x=\int_{0}^{\frac{\pi}{3}}\sec x\left ( \sec^{2}x \right )\textrm{d}x$ $=\int_{0}^{\frac{\pi}{3}}\sec x\left ( 1+ \tan^{2}x \right )\textrm{d}x$ $=\int_{0}^{\frac{\pi}{3}}\sec x + \sec x\tan^{2}x \textrm{d}x$ $=\int_{0}^{\frac{\pi}{3}}\sec x\textrm{d}x + \int_{0}^{\frac{\pi}{3}}\sec x\tan^{2}x \textrm{d}x$ $=\left [ \ln \left | \sec x + \tan x \right | \right ]_{0}^{\frac{\pi}{3}} + \int_{0}^{\frac{\pi}{3}}\sec x\tan^{2}x \textrm{d}x$ Using integration by parts to solve the second integral where, $u=\tan x\Rightarrow \frac{du}{dx}=\sec^{2}x$ $\frac{dv}{dx}=\sec x\tan x \Rightarrow v=\sec x$ $\int_{0}^{\frac{\pi}{3}}\sec^{3}x\textrm{d}x=\left [ \ln \left | \sec x + \tan x \right | \right ]_{0}^{\frac{\pi}{3}} +\left [ \sec x \tan x\right ]_{0}^{\frac{\pi}{3}} - \int_{0}^{\frac{\pi}{3}}\sec x\sec^{2}x \textrm{d}x$ Adding the integral to both sides gives, $2\int_{0}^{\frac{\pi}{3}}\sec^{3}x\textrm{d}x=\left [ \ln \left | \sec x + \tan x \right | \right ]_{0}^{\frac{\pi}{3}} +\left [ \sec x \tan x\right ]_{0}^{\frac{\pi}{3}}$ $\int_{0}^{\frac{\pi}{3}}\sec^{3}x\textrm{d}x=\frac{1}{2}\left [ \ln \left | \sec x + \tan x \right | + \sec x \tan x\right ]_{0}^{\frac{\pi}{3}}$ $=\frac{1}{2}\left \{ \left (\ln \left | \sec \frac{\pi}{3} + \tan \frac{\pi}{3} \right | + \sec \frac{\pi}{3} \tan \frac{\pi}{3} \right )-\left ( \ln \left | \sec 0 + \tan 0 \right | + \sec 0 \tan 0 \right )\right \}$ $=\frac{1}{2}\left \{ \left (\ln \left ( 2 + \sqrt{3} \right ) + 2\sqrt{3} \right )-\left ( 0 \right )\right \}$ $=\frac{1}{2}\ln \left ( 2 + \sqrt{3} \right ) + \sqrt{3}$