Answer
$$\int_0^{\pi/6}\frac{\sin t}{\cos^2t}dt=\frac{2}{\sqrt3}-1$$
Work Step by Step
To evaluate the integral $$\int_0^{\pi/6}\frac{\sin t}{\cos^2t}dt$$ we will use substitution $\cos t=z$ which gives us $-\sin tdt=dx\Rightarrow \sin tdt=-dz$ and the integration bounds would be: for $t=0$ we have $z=1$ and for $t=\pi/6$ we have $z=\sqrt 3/2$ so we get:
$$\int_0^{\pi/6}\frac{\sin t}{\cos^2t}dt=\int_1^{\sqrt3/2}-\frac{dz}{z^2}=-\left.\frac{z^{-1}}{-1}\right|_{-1}^{\sqrt3/2}=\left.\frac{1}{z}\right|_{-1}^{\sqrt3/2}=\frac{1}{\sqrt3/2}-\frac{1}{1}=\frac{2}{\sqrt3}-1$$
