Evaluate the definite integral _0^1 d x 1-x^2 - Vidyadip

Question

Evaluate the definite integral $\int_0^1 \frac{d x}{\sqrt{1-x^2}}$

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Answer

According to the question , $I = \int _ { 0 } ^ { 1 } \frac { 1 } { \sqrt { 1 - x ^ { 2 } } } d x$
= $\left[ \sin ^ { - 1 } x \right] _ { 0 } ^ { 1 } \quad \left[ \because \int_b^a \frac { 1 } { \sqrt { 1 - x ^ { 2 } } } d x = \sin ^ { - 1 } (a) - \sin ^ { - 1 } (b) \right]$
$= sin^{-1}(1)-sin^{-1}(0)$
$= \sin ^ { - 1 } \left( \sin \frac { \pi } { 2 } \right) - \sin ^ { - 1 } ( \sin 0 ) $

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

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