Question

\[\int\limits_0^a \sin^{- 1} \sqrt{\frac{x}{a + x}} dx\]

Answer 1

\[Let\, I = \int\limits_0^a \sin^{- 1} \sqrt{\frac{x}{a + x}} dx\]

\[Let, x = a \tan^2 \theta \Rightarrow \theta = \tan^{- 1} \sqrt{\frac{x}{a}}\]

\[When, x \to x ; \theta \to 0\ and\ x\ \to a ; \theta \to \frac{\pi}{4}\]

\[and\ dx\ = 2a \tan\theta se c^2 \theta d\theta\]

\[Then, \]

\[I = \int\limits_0^\frac{\pi}{4} \sin^{- 1} \sqrt{\frac{a \tan^2 \theta}{a + a \tan^2 \theta}} 2a \tan\theta se c^2 \theta\ d\theta\]

\[ \Rightarrow I = {2a \int}^\frac{\pi}{4}_0 \sin^{- 1} \left( \sin\theta \right) \tan\theta se c^2 \theta d\theta\]

\[ \Rightarrow I = {2a \int}^\frac{\pi}{4}_0 \theta \tan\theta se c^2 \theta d\theta\]

\[Let, \tan \theta = t \Rightarrow \theta = \tan^{- 1} t\]

\[ \Rightarrow se c^2 \theta d\theta = dt\]

\[when, \theta \to 0 ; t \to 0 and \theta \to \frac{\pi}{4} ; t \to 1\]

\[Then, I = 2a \int_0^1 \tan^{- 1} t\ t \ dt\]

\[ = 2a \int_0^1 \tan^{- 1} t\ t\ dt\]

\[ = 2a \left[ \tan^{- 1} t \frac{t^2}{2} \right]_0^1 - \frac{2a}{2} \int_0^1 \frac{t^2}{1 + t^2} dt\]

\[ = 2a\left[ \frac{\pi}{4} \times \frac{1}{2} - 0 \right] - a \int_0^1 \left[ 1 - \frac{1}{1 + t^2} \right] dt\]

\[ = 2a\left[ \frac{\pi}{8} \right] - a \left[ t - \tan^{- 1} t \right]_0^1 \]

\[ = \frac{\pi a}{4} - a\left[ 1 - \frac{\pi}{4} \right]\]

\[ = \frac{\pi a}{4} - a + \frac{\pi a}{4}\]

\[ = \frac{\pi a}{2} - a\]

\[ = a\left( \frac{\pi}{2} - 1 \right)\]