Answer
$\frac{2\sqrt{3}}{3}-1$
Work Step by Step
$\int_{0}^{\pi/6}\frac{sin~t}{cos^2~t}~dt$
Let $u = cos~t$
$\frac{du}{dt} = -sin~t$
$dt = -\frac{du}{sin~t}$
When $t = 0$, then $u = 1$
When $t = \frac{\pi}{6}$, then $u = \frac{\sqrt{3}}{2}$
$\int_{1}^{\sqrt{3}/2}\frac{sin~t}{u^2}\cdot (-\frac{du}{sin~t})$
$\int_{1}^{\sqrt{3}/2}-\frac{du}{u^2}$
$= \frac{1}{u}~\vert_{1}^{\sqrt{3}/2}$
$= \frac{2}{\sqrt{3}}-\frac{1}{1}$
$= \frac{2\sqrt{3}}{3}-1$
