Integrals — Class 12 Maths Solution

Let $I = \int {\frac{{{{\sin }^{ - 1}}x}}{{{{\left( {1 - {x^2}} \right)}^{3/4}}}}} dx = \int {\frac{{{{\sin }^{ - 1}}x}}{{\left( {1 - {x^2}} \right)\sqrt {1 - {x^2}} }}} dx$

Let's put ${\sin ^{ - 1}}x = t \Rightarrow \frac{1}{{\sqrt {1 - {x^2}} }}dx = dt$

and $x = \sin t \Rightarrow 1 - {x^2} = {\cos ^2}t$
$\Rightarrow$ $\cos t = \sqrt {1 - {x^2}}$

therefore,$I = \int {\frac{t}{{{{\cos }^2}t}}} dt = \int t \cdot {\sec ^2}tdt$

$= t \cdot \int {{{\sec }^2}} tdt - \int {\left( {\frac{d}{{dt}}t \cdot \int {{{\sec }^2}} tdt} \right)} dt$

$= t \cdot \tan t - \int 1 \cdot \tan tdt$

$= {\sin ^{ - 1}}x \cdot \frac{x}{{\sqrt {1 - {x^2}} }} + \log \left| {\sqrt {1 - {x^2}} } \right| + C$