Integral Calculus Question 235
Question: $ \int_{{}}^{{}}{\sqrt{\frac{x}{a^{3}-x^{3}}}\ dx=} $
Options:
A) $ {{\sin }^{-1}}{{( \frac{x}{a} )}^{3/2}}+c $
B) $ \frac{2}{3}{{\sin }^{-1}}{{( \frac{x}{a} )}^{3/2}}+c $
C) $ \frac{3}{2}{{\sin }^{-1}}{{( \frac{x}{a} )}^{3/2}}+c $
D) $ \frac{3}{2}{{\sin }^{-1}}{{( \frac{x}{a} )}^{2/3}}+c $
Show Answer
Answer:
Correct Answer: B
Solution:
Put $ x=a{{(\sin \theta )}^{2/3}}\Rightarrow dx=\frac{2}{3}a{{(\sin \theta )}^{-1/3}}\cos \theta ,d\theta $ \ $ \int_{{}}^{{}}{\sqrt{\frac{x}{a^{3}-x^{3}}},dx}=\int_{{}}^{{}}{\frac{{a^{1/2}}{{(\sin \theta )}^{1/3}}\frac{2}{3}a,{{(\sin \theta )}^{-1/3}}\cos \theta }{\sqrt{a^{3}-a^{3}{{\sin }^{2}}\theta }}},d\theta $ $ =\frac{2}{3}{a^{3/2}}\int_{{}}^{{}}{\frac{\cos \theta ,d\theta }{{a^{3/2}}\sqrt{1-{{\sin }^{2}}\theta }}}=\frac{2}{3}{{\sin }^{-1}}{{( \frac{x}{a} )}^{3/2}}+c $ .
BETA
