Mathematics

$$\int \sqrt { 1 - \sin  x } d x =$$


ANSWER

$$2\sqrt {1+\sin x}+C$$


SOLUTION
$$\\I=\int\sqrt{1-sinx}dx\\=\int\sqrt{1-sinx\times(\frac{1+sinx}{1+sinx})}dx$$
$$\\=\int\sqrt{(\frac{1^2-sin^2x}{1+sinx})}dx\\=\int(\frac{cosx}{\sqrt{1+sinx}})dx$$
let 1+sinx=t
then cosx dx=dt
$$\\or\>I=\int(\frac{1}{\sqrt t})dt$$
$$\\=(\frac{t^(\frac{1}{2})}{(\frac{1}{2})})+C\\2\sqrt{1+sinx}+C$$
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Single Correct Medium Published on 17th 09, 2020
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