Question

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

Answer 1

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

$={\int }^{\sqrt{2}}\sin \left(\frac{2x-\pi }{4}\right)dx$

$=-2^{\frac{3}{2}}\cos \left(u\right)$

$=-2^{\frac{3}{2}}\cos \left(\frac{2x-\pi }{4}\right)$

$=\sqrt{2}{\int }^{\sqrt{1-\sin \left(x\right)}}dx$

$=\sqrt{2}{\int }^{\sqrt{1-\sin \left(x\right)}}dx$

$=-4\cos \left(\frac{2x-\pi }{4}\right)+C$

$\sqrt{2}{\int }_0^{2\pi }\sqrt{1-\sin x}dx=8$