What is the integration of \[\int {\frac{{{{\sin }^{ – 1}}x}}{{{{\left( {1 – {x^2}} \right)}^{3/4}}}}} dx\] ~~~~~[NCERT Exemp. Q. 21,Page 164 ]

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


\[ = t \cdot \int {{{\sec }^2}} tdt – \int {\left( {\frac{d}{{dt}}t \cdot \int {{{\sec }^2}} tdt} \right)} dt\]


\[ = t \cdot \tan t – \int 1 \cdot \tan tdt\]

 

\[= {\sin ^{ – 1}}x \cdot \frac{x}{{\sqrt {1 – {x^2}} }} + \log \left| {\sqrt {1 – {x^2}} } \right| + C\]

 

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