Evaluate the definite integral in Exercise: _ 0 ^ 1 dx 1-x^ 2 - Vidyadip

Question

Evaluate the definite integral in Exercise:
$\int\limits_{0}^{1}\frac{\text{dx}}{\sqrt{1-\text{x}^{2}}}$

✓

Answer

$\text{Let}\ \text{I}=\int\limits_{0}^{1}\frac{\text{dx}}{\sqrt{1-\text{x}^{2}}}$

$\int\frac{\text{dx}}{\sqrt{1-\text{x}^{2}}}=\sin^{-1}\text{x}=\text{F}\text{(x)}$

By second fundamental theorem of calculus, we obtain

$\text{I}=\text{F}(1)-\text{F}(0)$

$=\sin^{-1}(1)-\sin^{-1}(0)$

$=\frac{\pi}{2}-0$

$=\frac{\pi}{2}$

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