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Home  >>  JEEMAIN and NEET  >>  Mathematics  >>  Class12  >>  Integral Calculus

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.
answered Jan 2, 2014 by meena.p
 
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