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
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