Question

Solve:
${\int }_0^{2\pi }\sqrt{1-\sin x}dx$

Answer 1

${\int }_0^{2\pi }\sqrt{1-sinx}.dx$

$={\int }_0^{2\pi }\sqrt{1-2sin\frac{x}{2}.cos\frac{x}{2}.dx}$

$={\int }_0^{2\pi }\sqrt{si{n}^2\frac{x}{2}+co{s}^2\frac{x}{2}-2sin\frac{x}{2}.cos\frac{x}{2}.dx}$

$={\int }_0^{2\pi }\sqrt{(sin\frac{x}{2}+cos\frac{x}{2}{)}^2}.dx$

$={\int }_0^{2\pi }(sin\frac{x}{2}+cos\frac{x}{2})dx$

$={[-\frac{cos\frac{x}{2}}{1/2}-\frac{sin\frac{x}{2}}{1/2}]}_0^{2\pi }$

$=[\frac{-cos\pi }{1/2}-\frac{sin\pi }{1/2}]-[\frac{-cos0}{1/2}-\frac{-sin0}{1/2}]$

$=[-2(-1)-0]-[-2(1)-0]$

$=2+2$

$=4$