Question
Evaluate $\int\frac{1}{\sqrt{1-\text{x}^2}}\text{ dx}$

Answer

Let $\text{I}=\int\frac{\text{dx}}{\sqrt{1-\text{x}^2}}$
Let $\text{x}=\sin\theta$
$\text{dx}=\cos\theta$
$\therefore\ \text{I}=\int\frac{\cos\theta}{\cos\theta}\text{ d}\theta$
$=\int\text{d}\theta$
$=\theta+\text{C}$
$=\sin^{-1}\text{x}+\text{C}$ $(\because\text{ x}=\sin\theta)$

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