How to integrate $\int {{{\sin }^{ – 1}}} \sqrt {\frac{x}{{a + x}}} dx$ ~~~~~[NCERT Exemp. Q. 40,Page 166 ]

Let $I = \int {{{\sin }^{ – 1}}} \sqrt {\frac{x}{{a + x}}} dx$


Let’s put $x = a{\tan ^2}\theta $


$ \Rightarrow $$dx = 2a\tan \theta {\sec ^2}\theta d\theta $


therefore,$I = \int {{{\sin }^{ – 1}}} \sqrt {\frac{{a{{\tan }^2}\theta }}{{a + a{{\tan }^2}\theta }}} \left( {2a\tan \theta \cdot {{\sec }^2}\theta } \right)d\theta $


$ = 2a\int {{{\sin }^{ – 1}}} \left( {\frac{{\tan \theta }}{{\sec \theta }}} \right)\tan \theta \cdot {\sec ^2}\theta d\theta $


$ = 2a\int {{{\sin }^{ – 1}}} (\sin \theta )\tan \theta \cdot {\sec ^2}\theta d\theta $


$ = \mathop {2a\int \theta \cdot \tan \theta {{\sec }^2}\theta d\theta }\limits_{{\rm{I}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{II}}} $


$\quad $ I $\quad $ II
$ = 2a\left[ {\theta \cdot \int {\tan } \theta \cdot {{\sec }^2}\theta d\theta – \int {\left( {\frac{d}{{d\theta }}\theta \cdot \int {\tan } \theta \cdot {{\sec }^2}\theta d\theta } \right)} d\theta } \right]$


$\left[ {\begin{array}{*{20}{l}}{{\rm{ Let’s put }}\,\,\,\,\,\,\,\,\,\,\tan \theta = t}\\{ \Rightarrow \sec \theta \cdot \tan \theta \cdot d\theta = dt}\\{ \Rightarrow \int {\tan } \theta {{\sec }^2}\theta d\theta = \int t dt}\end{array}} \right]$


$ = 2a\left[ {\theta \cdot \frac{{{{\tan }^2}\theta }}{2} – \int {\frac{{{{\tan }^2}\theta }}{2}} d\theta } \right]$


$ = a\theta {\tan ^2}\theta – a\int {\left( {{{\sec }^2}\theta – 1} \right)} d\theta $


$ = a\theta \cdot {\tan ^2}\theta – a\tan \theta + a\theta + C$


$ = a\left[ {\frac{x}{a}{{\tan }^{ – 1}}\sqrt {\frac{x}{a}} + {{\tan }^{ – 1}}\sqrt {\frac{x}{a}} } \right] + C$

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