Integrals — Class 12 Maths Solution

Let $I = \int {\cfrac{{x{{\cos }^{ - 1}}x}}{{\sqrt {1 - {x^2}} }}} dx$
Put ${\cos ^{ - 1}}x = t$ $\Rightarrow$ $\cfrac{{ - 1}}{{\sqrt {1 - {x^2}} }}dx = dt$

$\therefore$ $I = - \int {t\cos tdt} = - \left[ {t\int {\cos t\,dt} - \int {\left( {\cfrac{d}{{dt}}\left( t \right)\int {\cos t} \,dt} \right)} dt} \right]$

$= - t\sin t + \int {\sin t\,dt} = - t\sin t - \cos t + C$

$= - t\sqrt {1 - {{\cos }^2}t} - \cos t + C = - {\cos ^{ - 1}}x\sqrt {1 - {x^2}} - x + C$

$= - \left[ {\sqrt {1 - {x^2}} {{\cos }^{ - 1}}x + x} \right] + C$