Mathematics

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


SOLUTION
$$\displaystyle\int{\sqrt{1-{x}^{2}}dx}$$
Let $$x=\sin{\theta}\Rightarrow dx=\cos{\theta}d\theta$$
$$=\displaystyle\int{\sqrt{1-{\sin}^{2}{\theta}}\cos{\theta}d\theta}$$
$$=\displaystyle\int{\sqrt{{\cos}^{2}{\theta}}\cos{\theta}d\theta}$$
$$=\displaystyle\int{{\cos}^{2}{\theta}d\theta}$$
$$=\displaystyle\int{{\cos}^{2}{\theta}d\theta}$$ since $$\cos{2\theta}=2{\cos}^{2}{\theta}-1\Rightarrow {\cos}^{2}{\theta}=\dfrac{1}{2}\left(1+\cos{2\theta}\right)$$
$$=\dfrac{1}{2}\displaystyle\int{\left(1+\cos{2\theta}\right)d\theta}$$
$$=\dfrac{1}{2}\left[\theta+\dfrac{\sin{2\theta}}{2}\right]+c$$ 
$$=\dfrac{{\sin}^{-1}{x}}{2}+\dfrac{\sin{2\left({\sin}^{-1}{x}\right)}}{4}+c$$ where $$\theta={\sin}^{-1}{x}$$
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Subjective Medium Published on 17th 09, 2020
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