Integral Calculus Question 108
Question: $ \int_{{}}^{{}}{\frac{x^{2}}{{{(9-x^{2})}^{3/2}}}\ dx=} $
Options:
A) $ \frac{x}{\sqrt{9-x^{2}}}-{{\sin }^{-1}}\frac{x}{3}+c $
B) $ \frac{x}{\sqrt{9-x^{2}}}+{{\sin }^{-1}}\frac{x}{3}+c $
C) $ {{\sin }^{-1}}\frac{x}{3}-\frac{x}{\sqrt{9-x^{2}}}+c $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
Put $ x=3\sin \theta \Rightarrow dx=3\cos \theta ,d\theta , $ therefore
$ \int_{{}}^{{}}{\frac{x^{2}}{{{(9-x^{2})}^{3/2}}},dx}=\int_{{}}^{{}}{\frac{9{{\sin }^{2}}\theta }{{{(9-9{{\sin }^{2}}\theta )}^{3/2}}.,3\cos \theta },d\theta } $
$ =\int_{{}}^{{}}{\frac{27{{\sin }^{2}}\theta \cos \theta }{27{{\cos }^{3}}\theta }},d\theta =\int_{{}}^{{}}{{{\tan }^{2}}\theta ,d\theta }=\int_{{}}^{{}}{({{\sec}^{2}}\theta -1),d\theta } $
$ =\tan \theta -\theta +c=\tan { {{\sin }^{-1}}( \frac{x}{3} ) }-{{\sin }^{-1}}( \frac{x}{3} )+c $
$ =\tan {{\tan }^{-1}}( \frac{( \frac{x}{3} )}{\sqrt{1-(x^{2}/9)}} )-{{\sin }^{-1}}( \frac{x}{3} )+c $
$ =\frac{x}{\sqrt{9-x^{2}}}-{{\sin }^{-1}}( \frac{x}{3} )+c. $
BETA
