Originally Posted by

**Hellbent** Hi,

$\displaystyle \text{Show that}\,\,\int_0^\frac{\pi}{6} \frac{1}{2\cos^2\,\theta-1}=\frac{1}{2}\ln(2 + \sqrt3)$

$\displaystyle \int \frac{1}{2\cos^2 \theta - 1} = \frac{1}{2} (\log(\sin(\theta) + \cos(\theta)) - \log(\cos(\theta) - \sin(\theta))) + K$

$\displaystyle \frac{1}{2} (\log(\sin(\theta) + \cos(\theta)) - \log(\cos(\theta) - \sin(\theta))) + K$

$\displaystyle F|_a^b\Rightarrow F |_0^\frac{\pi}{6}$

$\displaystyle = \frac{1}{2} \bigg(\log\bigg(\sin\bigg(\frac{\pi}{6}\bigg) + \cos\bigg(\frac{\pi}{6}\bigg)\bigg) - \log\bigg(\cos\bigg(\frac{\pi}{6}\bigg) - \sin\bigg(\frac{\pi}{6}\bigg)\bigg)\bigg)\bigg -

\frac{1}{2} \bigg(\log\bigg(\sin\bigg(0\bigg) + \cos\bigg(0\bigg)\bigg) - \log\bigg(\cos\bigg(0\bigg) - \sin\bigg(0\bigg)\bigg)\bigg)$

$\displaystyle = \frac{1}{2}\log\bigg(\frac{1}{2} + \frac{\sqrt3}{2} - \log\bigg(\frac{\sqrt3}{2} - \frac{1}{2}\bigg)\bigg)$