\( 1 \leftarrow \int _ { 0 } ^ { \pi / 6 } 3 \sec ^ { 2 } x d x \)

To solve the integral \( \int _ { 0 } ^ { \pi / 6 } 3 \sec ^ { 2 } x \, dx \), we start by recognizing that the derivative of \( \tan x \) is \( \sec^2 x \). Therefore, we can rewrite the integral: \[ \int 3 \sec^2 x \, dx = 3 \tan x + C \] Now we evaluate at the limits \( 0 \) to \( \frac{\pi}{6} \): \[ \left[ 3 \tan x \right] _ { 0 } ^ { \pi / 6 } = 3 \tan \left( \frac{\pi}{6} \right) - 3 \tan(0) \] Since \( \tan(0) = 0 \) and \( \tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}} \): \[ = 3 \left( \frac{1}{\sqrt{3}} \right) - 0 = \frac{3}{\sqrt{3}} = \sqrt{3} \] So, the final value of the integral is: \[ \int _ { 0 } ^ { \pi / 6 } 3 \sec ^ { 2 } x \, dx = \sqrt{3} \]