What is the integration of \[\int {\sqrt {\frac{{a + x}}{{a – x}}} } dx\] ~~~~~[NCERT Exemp. Q. 11,Page 164 ]

Let \[I = \int {\sqrt {\frac{{a + x}}{{a – x}}} } dx\]

Let’s put \[ x = {\mathop{\rm acos}\nolimits} 2\theta \]

\[ \Rightarrow dx = – a \cdot \sin 2\theta \cdot 2 \cdot d\theta \]

therefore,\[ I = – 2\int {\sqrt {\frac{{a + a\cos 2\theta }}{{a – a\cos 2\theta }}} } \cdot a\sin 2\theta d\theta \]


\[ = – 2a\int {\sqrt {\frac{{1 + \cos 2\theta }}{{1 – \cos 2\theta }}} } \sin 2\theta d\theta = – 2a\int {\sqrt {\frac{{2{{\cos }^2}\theta }}{{2{{\sin }^2}\theta }}} } \sin 2\theta d\theta \]

\[ = – 2a\int {\cot } \theta \cdot \sin 2\theta d\theta = – 2a\int {\frac{{\cos \theta }}{{\sin \theta }}} \cdot 2\sin \theta \cdot \cos \theta d\theta \]

\[ = – 4a\int {{{\cos }^2}} \theta d\theta = – 2a\int {(1 + \cos 2\theta )} d\theta \]

\[ = – 2a\left[ {\theta + \frac{{\sin 2\theta }}{2}} \right] + C\]

\[ = – 2a\left[ {\frac{1}{2}{{\cos }^{ – 1}}\frac{x}{a} + \frac{1}{2}\sqrt {1 – \frac{{{x^2}}}{{{a^2}}}} } \right] + C\]

\[ = – a\left[ {{{\cos }^{ – 1}}\left( {\frac{x}{a}} \right) + \sqrt {1 – \frac{{{x^2}}}{{{a^2}}}} } \right] + C\]

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