Answer
$\displaystyle \frac{1}{2}$
Work Step by Step
$\displaystyle \int_{0}^{π/3}\frac{\sin(\theta)+\sin(\theta)\tan^2(\theta)}{\sec^2(\theta)} d\theta$
We can manipulate the fraction to make it easier to integrate.
$\displaystyle \int_{0}^{π/3}\frac{\sin(\theta)+\sin(\theta)[\frac{\sin^2(\theta)}{\cos^2(\theta)}]}{\frac{1}{\cos^2(\theta)}} d\theta$
$\displaystyle \int_{0}^{π/3} \cos^2(\theta)[\sin(\theta)+\sin(\theta)[\frac{\sin^2(\theta)}{\cos^2(\theta)}]]d\theta$
$\displaystyle \int_{0}^{π/3} \sin(\theta)\cos^2(\theta)+\sin^3(\theta)d\theta$
Use the trigonometric identity $\quad\sin^2(\theta)+\cos^2(\theta)=1\quad\rightarrow\quad \sin^2(\theta)=1-\cos^2(\theta)$
$\displaystyle \int_{0}^{π/3} \sin(\theta)\cos^2(\theta)+\sin(\theta)(1-\cos^2(\theta))d\theta$
$\displaystyle \int_{0}^{π/3} \sin(\theta)\cos^2(\theta)+\sin(\theta)-\sin(\theta)\cos^2(\theta)d\theta$
The two $\quad\sin(\theta)\cos^2(\theta)\quad$ cancel out and we are left with:
$\displaystyle \int_{0}^{π/3} \sin(\theta)d\theta$
$-\cos(\theta)|_0^{π/3}$
$\displaystyle -\cos(\frac{π}{3})-[-\cos(0)]$
$\displaystyle -\frac{1}{2}+1$
$\displaystyle \frac{1}{2}$
