To solve \(\int x \sec^2 x \, dx\), we use integration by parts. Let \(u = x\) and \(dv = \sec^2 x \, dx\). Then \(du = dx\) and \(v = \tan x\).
Applying integration by parts:
\(\int x \sec^2 x \, dx = x \tan x - \int \tan x \, dx\)
The integral \(\int \tan x \, dx\) is \(-\ln |\cos x|\).
Thus,
\(\int x \sec^2 x \, dx = x \tan x + \ln |\cos x|\)
Now, evaluate from \(x = \frac{\pi}{6}\) to \(x = \frac{\pi}{3}\):
\(\left[ x \tan x + \ln |\cos x| \right]_{\frac{\pi}{6}}^{\frac{\pi}{3}}\)
Calculate at \(x = \frac{\pi}{3}\):
\(\frac{\pi}{3} \tan \frac{\pi}{3} + \ln |\cos \frac{\pi}{3}| = \frac{\pi}{3} \sqrt{3} + \ln \frac{1}{2}\)
Calculate at \(x = \frac{\pi}{6}\):
\(\frac{\pi}{6} \tan \frac{\pi}{6} + \ln |\cos \frac{\pi}{6}| = \frac{\pi}{6} \frac{1}{\sqrt{3}} + \ln \frac{\sqrt{3}}{2}\)
Subtract the two results:
\(\left( \frac{\pi}{3} \sqrt{3} + \ln \frac{1}{2} \right) - \left( \frac{\pi}{6} \frac{1}{\sqrt{3}} + \ln \frac{\sqrt{3}}{2} \right)\)
Simplify using logarithm properties:
\(\frac{\pi}{3} \sqrt{3} - \frac{\pi}{6} \frac{1}{\sqrt{3}} + \ln \frac{1}{2} - \ln \frac{\sqrt{3}}{2}\)
\(= \frac{5}{18} \sqrt{3} \pi - \frac{1}{2} \ln 3\)
