Mathematics

$$\int \sqrt {1 + \sin x}dx =$$


ANSWER

$$-2\sqrt {1 - \sin x} + c$$


SOLUTION
$$\displaystyle\int \sqrt{1+\sin x} dx$$
$$=\displaystyle\int \dfrac{\sqrt{1+\sin x}.\sqrt{1-\sin x}}{\sqrt{1-\sin x}}\ dx$$
$$=\displaystyle\int \dfrac{\sqrt{(1+\sin x)(1-\sin x)}}{\sqrt{1-\sin x}}dx$$
$$=\displaystyle\int\dfrac{\sqrt{1-\sin^{2}x}}{\sqrt{1-\sin x}}dx=\int \dfrac{\cos x}{\sqrt{1-\sin x}}dx$$
Let $$u=\sqrt{1-\sin x}$$
$$du=\dfrac{1}{\sqrt{1-\sin x}}(-\cos x)dx$$
$$-2du=\dfrac{\cos x}{2\sqrt{1-\sin x}}dx$$
$$=\displaystyle\int -2du=-2u+c$$
$$=-2\sqrt{1-\sin x}+c$$
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Single Correct Medium Published on 17th 09, 2020
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