Answer
$$ \frac{\sqrt{3}}{3}$$
Work Step by Step
Given
$$ \int_{0}^{\pi / 6} \sec ^{2}\left(2 x-\frac{\pi}{6}\right) d x$$
Let
$$ u= 2 x -\frac{\pi}{6} \ \ \ \Rightarrow \ \ \ du =2dx$$
At $$ x= 0\to u= -\pi / 6, \ \ x= \pi/6\to u= \pi / 6$$
Then
\begin{aligned} \int_{0}^{\pi / 6} \sec ^{2}\left(2 x-\frac{\pi}{6}\right) d x&=
\frac{1}{2} \int_{-\pi / 6}^{\pi / 6} \sec ^{2} u d u\\
&=\left.\frac{1}{2} \tan u\right|_{-\pi / 6} ^{\pi / 6}\\
&=\frac{1}{2}\left( \frac{\sqrt{3}}{3}+ \frac{\sqrt{3}}{3} \right)\\
&= \frac{\sqrt{3}}{3} \end{aligned}
