Mathematics

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

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

##### 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
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

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