Integral Calculus Question 245
Question: $ \int_{{}}^{{}}{\sqrt{\frac{a-x}{x}}\ dx=} $
Options:
A) $ a[ {{\sin }^{-1}}\sqrt{\frac{x}{a}}+\sqrt{\frac{x}{a}}\sqrt{\frac{a-x}{a}} ]+c $
B) $ {{\sin }^{-1}}\frac{x}{a}+\frac{x}{a}\sqrt{a^{2}-x^{2}}+c $
C) $ a[ {{\sin }^{-1}}\frac{x}{a}-\frac{x}{a}\sqrt{a^{2}-x^{2}} ]+c $
D) $ {{\sin }^{-1}}\frac{x}{a}-\frac{x}{a}\sqrt{a^{2}-x^{2}}+c $
Show Answer
Answer:
Correct Answer: A
Solution:
$ I=\int_{{}}^{{}}{\sqrt{\frac{a-x}{x}},dx} $ .
Put $ x=a{{\sin }^{2}}\theta \Rightarrow dx=2a\sin \theta \cos \theta ,d\theta , $ then
$ I=\int_{{}}^{{}}{\sqrt{\frac{{{\cos }^{2}}\theta }{{{\sin }^{2}}\theta }}},.,2a\sin \theta \cos \theta ,d\theta $
$ =a\int_{{}}^{{}}{2{{\cos }^{2}}\theta ,d\theta }=a\int_{{}}^{{}}{(1+\cos 2\theta ),d\theta } $
$ =a,[ {{\sin }^{-1}}\sqrt{\frac{x}{a}}+\sqrt{\frac{x}{a}},.,\sqrt{\frac{a-x}{a}} ]+c $ .
BETA
