# Integrate : $\int \cos^{-1} \sqrt {\large\frac{x}{a+x}}$$dx (a)\;\frac{\pi}{2} x -\frac{x}{2a} \sqrt {\frac{x}{a}}+\frac{1}{2} \sqrt {\frac{x}{a}}-\frac{1}{2} \tan ^{-1} (\sqrt{\frac{x}{a}})+c \\(b)\;\frac{\pi}{2} x -\frac{x}{2a} \sqrt {\frac{x}{a}}+\frac{1}{2} \sqrt {\frac{x}{a}}-\frac{1}{2} \cot ^{-1} (\sqrt{\frac{x}{a}})+c \\(c)\; \frac{\pi}{2} x -\frac{x}{2a} \sqrt {\frac{x}{a}}+\frac{1}{2} \sqrt {\frac{x}{a}}-\frac{1}{2} \sin ^{-1} (\sqrt{\frac{x}{a}})+c \\ (d)\;\frac{\pi}{2} x -\frac{x^2}{2a} \sqrt {\frac{x}{a}}+\frac{x}{2} \sqrt {\frac{x}{a}}-\frac{1}{7} \tan ^{-1} (\sqrt{\frac{x}{a}})+c ## 1 Answer \int \large\frac{\pi}{2}$$- \sin ^{-1} \sqrt {\large\frac{x}{a+x}}$$dx =>\int \large\frac{\pi}{2}$$- \int \sin ^{-1} \sqrt {\large\frac{x}{a+x}}$$dx => \large\frac{\pi}{2} -$$ \int \sin ^{-1} \sqrt {\large\frac{x}{a+x}}$$dx x= a \tan ^2 t differentiate with respect to x dx= 2a \tan t. \sec^2 t.dt => \large\frac{z}{2}$$x- \int (\sin ^{-1} (\sin t) .2a . \tan t . \sec^2 t.dt$
=> $\large\frac{x \pi }{2} $$-\int t.2a. \tan t. \sec^2t.dt \tan t =y \sec^2 t.dt=dy => \large\frac{\pi}{2}$$-\int \tan ^{-1} (y) .y.dy$
=> $\large\frac{\pi}{2} $$x -\tan ^{-1} y.\large\frac{y^2}{2} +\large\frac{1}{2} \int \large\frac{y^2}{1+y^2}$$dy$
=> $\large\frac{\pi}{2} $$\large\frac{y^2}{2}$$ \tan^{-1}y +\large\frac{1}{2} \int dy -\large\frac{1}{2} \int \frac{1}{1+y^2}$$dy \large\frac{\pi}{2} x -\frac{x}{2a} \sqrt {\frac{x}{a}}+\frac{1}{2} \sqrt {\frac{x}{a}}-\frac{1}{2}$$\tan ^{-1} (\sqrt{\frac{x}{a}})+c$
Hence a is the correct answer.