Mathematics

# $\displaystyle\int{\dfrac{dx}{\sqrt{1-{x}^{2}}}}$

##### SOLUTION
Let $x=\sin{\theta}\Rightarrow\,dx=\cos{\theta}d\theta$
$1-{x}^{2}=1-{\sin}^{2}{\theta}={\cos}^{2}{\theta}$
$\sqrt{1-{x}^{2}}=\cos{\theta}$
$\displaystyle\int{\dfrac{dx}{\sqrt{1-{x}^{2}}}}$
$=\displaystyle\int{\dfrac{\cos{\theta}d\theta}{\cos{\theta}}}$
$=\displaystyle\int{d\theta}$
$=\theta+c$
$={\sin}^{-1}{x}+c$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 124

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