I=∫ sin-1xa+x dx Put x=a tan2 t dx=a 2 tan t.sec2t dt I=∫ sin-1 a tan2 ta+a tan2t (2a tan t sec2 t dt) =∫ sin-1 tan tsec t.2a tan t sec2t dt =∫ sin-1 (sin t).2a tan t.sec2t dt =(2a) ∫ t. (tan t.sec2t) dt =2attan2t2-∫ 1.tan2t2dt =2atan-1xax2a -12 (tan t-t) =tan-1xax-axa-tan-1xa =tan-1xa (x+a)-ax+c

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$\int S i n^{- 1} \sqrt{\frac{x}{a + x}} d x =$

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$\cos e c (\sqrt{\frac{x}{a + x}}) (\frac{x}{a}) + a x + c$

$C o s^{- 1} (\sqrt{\frac{x}{a}}) (a + x) - \sqrt{a x} + c$

$C o s^{- 1} (\sqrt{\frac{x}{a}}) {(a - x)}^{2} - \sqrt{a x} + c$

$T a n^{- 1} (\sqrt{\frac{x}{a}}) (a + x) - \sqrt{a x} + c$

answer is D.

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Detailed Solution

$I = \int \sin^{- 1} \sqrt{\frac{x}{a + x}} dx Put x = a \tan^{2} t d x = a 2 \tan t . \sec^{2} t dt I = \int \sin^{- 1} \sqrt{\frac{a \tan^{2} t}{a + a \tan^{2} t}} (2 a \tan t \sec^{2} t dt) = \int \sin^{- 1} (\frac{\tan t}{\sec t}) . 2 a \tan t \sec^{2} t dt = \int \sin^{- 1} (\sin t) . 2 a \tan t . \sec^{2} t dt = (2 a) \int t . (\tan t . \sec^{2} t) dt = 2 a [t (\frac{\tan^{2} t}{2}) - \int 1 . \frac{\tan^{2} t}{2} dt] = 2 a [(\tan^{- 1} \sqrt{\frac{x}{a}}) \frac{x}{2 a} \frac{- 1}{2} (\tan t - t)] = (\tan^{- 1} \sqrt{\frac{x}{a}}) x - a (\sqrt{\frac{x}{a}} - \tan^{- 1} \sqrt{\frac{x}{a}}) = (\tan^{- 1} \sqrt{\frac{x}{a}}) (x + a) - \sqrt{ax} + c$