$\cfrac{{{{\sin }^{ - 1}}x}}{{\sqrt {1 - {x^2}} }}$
$\cfrac{{{{\sin }^{ - 1}}x}}{{\sqrt {1 - {x^2}} }}$
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NCERT & Exemplar
Let$I = \int {\cfrac{{{{\sin }^{ - 1}}x}}{{\sqrt {1 - {x^2}} }}dx}$
Put ${\sin ^{ - 1}}x = t$ $\Rightarrow$ $\cfrac{1}{{\sqrt {1 - {x^2}} }}dx = dt$
$\Rightarrow$ $I = \int t dt = \cfrac{1}{2}{t^2} + C = \cfrac{1}{2}{\left( {{{\sin }^{ - 1}}x} \right)^2} + C$
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