Mathematics

Integrate $$\displaystyle \int _{ 0 }^{ 2\pi  }{ \sqrt {1+\sin x }}dx$$


ANSWER

$$4$$


SOLUTION
$$\int_0^{2\pi } {\sqrt {1 + \sin x} } dx$$
$$ = \int_0^{2\pi } {\sqrt {{{\sin }^2}\frac{x}{2} + {{\cos }^2}\frac{x}{2} + 2\sin \frac{x}{2}\cos } \frac{x}{2}} dx$$    ($${\therefore {{\sin }^2}x + {{\cos }^2}x = 1}$$ and $$\sin 2x = 2\sin x\cos x$$)
$$= \int_0^{2\pi } {\sqrt {{{\left( {\sin \frac{x}{2} + \cos \frac{x}{2}} \right)}^2}} dx} $$
$$ = \int_0^{2\pi } {\left( {\sin \frac{x}{2} + \cos \frac{x}{2}} \right)} dx$$
$$ = \left[ { - 2\cos \frac{x}{2} + 2\sin \frac{x}{2}} \right]_0^{2\pi }$$
$$ = 2\left[ {\sin \frac{x}{2} - \cos \frac{x}{2}} \right]_0^{2\pi }$$
$$ = 2\left[ {1 + 1} \right]$$
$$ = 4$$
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Single Correct Medium Published on 17th 09, 2020
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