Integrals — Class 12 Maths Solution

Let $I = \int {x{{\sin }^{ - 1}}x\,dx} = \int {{{\sin }^{ - 1}}x \cdot x\,dx}$

$= {\sin ^{ - 1}}x \cdot \left( {\cfrac{{{x^2}}}{2}} \right) - \int {\left( {\cfrac{d}{{dx}}\left( {{{\sin }^{ - 1}}x} \right) \cdot \cfrac{{{x^2}}}{2}} \right)dx}$

$= {\sin ^{ - 1}}x \cdot \left( {\cfrac{{{x^2}}}{2}} \right) - \int {\cfrac{1}{{\sqrt {1 - {x^2}} }} \cdot \cfrac{{{x^2}}}{2}dx}$

$= \cfrac{{{x^2}}}{2}{\sin ^{ - 1}}x - \cfrac{1}{2}\int {\cfrac{{{x^2}}}{{\sqrt {1 - {x^2}} }}dx} = \cfrac{{{x^2}}}{2}{\sin ^{ - 1}}x - \cfrac{1}{2}{I_1}$

$\Rightarrow$ $I = \cfrac{{{x^2}}}{2}{\sin ^{ - 1}}x - \cfrac{1}{2}{I_1}$

Where, ${I_1} = \int {\cfrac{{{x^2}}}{{\sqrt {1 - {x^2}} }}dx}$

Put $x = \sin \theta$ $\Rightarrow$ $dx = \cos \theta d\theta$

$\therefore$ ${I_1} = \int {\cfrac{{{{\sin }^2}\theta }}{{\sqrt {1 - {{\sin }^2}\theta } }}\cos \theta d\theta = \int {\cfrac{{{{\sin }^2}\theta }}{{\cos \theta }}} \cos \theta d\theta }$

$= \int {{{\sin }^2}\theta d\theta } = \cfrac{1}{2}\int {\left( {1 - \cos 2\theta } \right)d\theta }$

$= \cfrac{1}{2}\int {d\theta } - \cfrac{1}{2}\int {\cos 2\theta d\theta } = \cfrac{1}{2}\theta - \cfrac{1}{2}\cfrac{{\sin 2\theta }}{2} + {C_1}$

$= \cfrac{1}{2}\theta - \cfrac{1}{2}\sin \theta \cos \theta + {C_1} = \cfrac{1}{2}{\sin ^{ - 1}}x - \cfrac{1}{2}x\sqrt {1 - {x^2}} + {C_1}$

…(ii)

From (i) and (ii),

we get
$I = \cfrac{{{x^2}}}{2}{\sin ^{ - 1}}x - \cfrac{1}{2}\left[ {\cfrac{1}{2}{{\sin }^{ - 1}}x - \cfrac{1}{2}x\sqrt {1 - {x^2}} } \right] + C$ $\left[ {C = \cfrac{{ - {C_1}}}{2}} \right]$

$= \cfrac{{\left( {2{x^2} - 1} \right)}}{4}{\sin ^{ - 1}}x + \cfrac{{x\sqrt {1 - {x^2}} }}{4} + C$