Calculate various definite integrals of sine

Compute the integral

$\int_a^b \sin x \, dx$

for the following values of $a,b$ :

$\displaystyle{\int_0^{\frac{\pi}{6}} \sin x \, dx = 1 - \cos \frac{\pi}{6} = 1 - \frac{\sqrt{3}}{2}}$ .
$\displaystyle{\int_0^{\frac{\pi}{4}} \sin x \, dx = 1 - \cos \frac{\pi}{4} = 1 - \frac{\sqrt{2}}{2}}$ .
$\displaystyle{\int_0^{\frac{\pi}{3}} \sin x \, dx = 1 - \cos \frac{\pi}{3} = 1 - \frac{1}{2} = \frac{1}{2}}$ .
$\displaystyle{\int_0^{\frac{\pi}{2}} \sin x \, dx = 1 - \cos \frac{\pi}{2} = 1$ .
$\displaystyle{\int_0^{\pi} \sin x \, dx = 1 - \cos \pi = 2$ .
$\displaystyle{\int_0^{2\pi} \sin x \, dx = 1 - \cos 2 \pi = 0}$ .
$\displaystyle{\int_{-1}^1 \sin x \, dx = -\cos 1 - \cos(-1) = 0}$ .
$\displaystyle{\int_{-\frac{\pi}{6}}^{\frac{\pi}{4}} \sin x \, dx = \cos \left( -\frac{\pi}{6} \right) - \cos \frac{\pi}{4} = \frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{2} = \frac{\sqrt{3} - \sqrt{2}}{2}}$ .