Integrals — Class 12 Maths Solution

: $\int\limits_0^1 {{{\sin }^{ - 1}}xdx} = \int\limits_0^1 {{{\sin }^{ - 1}}x \cdot 1dx}$

Integrating by parts
$= \left[ {{{\sin }^{ - 1}}x\int {dx} - \left( {\int {\cfrac{d}{{dx}}\left( {{{\sin }^{ - 1}}x} \right) \cdot \int {\left( 1 \right)} dx} } \right)dx} \right]_0^1$

$= \left[ {{{\sin }^{ - 1}}x \cdot x - \int {\cfrac{1}{{\sqrt {1 - {x^2}} }} \times x} dx} \right]_0^1$

$= \left[ {x{{\sin }^{ - 1}}x + \cfrac{1}{2}\int {\cfrac{{ - 2x}}{{\sqrt {1 - {x^2}} }}} dx} \right]_0^1 = \left[ {x{{\sin }^{ - 1}}x + \cfrac{1}{2}\left[ {\cfrac{{{{\left( {1 - {x^2}} \right)}^{1/2}}}}{{1/2}}} \right]} \right]_0^1$

$= \left[ {\left( {{{\sin }^{ - 1}}1} \right) + \cfrac{1}{2} \times \cfrac{2}{1}\left( {0 - 1} \right)} \right] = \cfrac{\pi }{2} - 1$