Originally Posted by

**ashura** Seek explanation of getting the second line, $\displaystyle = 8(\pi+ \frac{\pi}{2}) - (0)$.

$\displaystyle A=8 \left [\theta + 2sin\theta + \frac{\theta}{2} + \frac{sin2\theta}{4} \right ]^\pi_0$

$\displaystyle = 8(\pi+ \frac{\pi}{2}) - (0)$

$\displaystyle =8 \pi + 4 \pi = 12\pi $ Ans

I know that $\displaystyle sin2 \theta = 2sin \theta cos \theta$

so that,

$\displaystyle 2sin \theta + \frac{sin2 \theta}{4} = 2sin\theta + \frac{2sin \theta cos \theta}{4}$

$\displaystyle =2sin \theta \left ( 1 + \frac {cos \theta}{4} \right )$