Integral Calculus Question 137
Question: $ \int{\sqrt{\frac{x}{1-x}}dx} $ is equal to
Options:
A) $ {{\sin }^{-1}}\sqrt{x}+c $
B) $ {{\sin }^{-1}}{\sqrt{x}-\sqrt{x(1-x)}}+c $
C) $ {{\sin }^{-1}}\sqrt{x(1-x)}+c $
D) $ {{\sin }^{-1}}\sqrt{x}-\sqrt{x(1-x)}+c $
Show Answer
Answer:
Correct Answer: D
Solution:
[d] Put $ x={{\sin }^{2}}\theta \Rightarrow dx=2\sin \theta \cos \theta $
$ \therefore \int{\sqrt{\frac{x}{1-x}dx}}=\int{\frac{\sin \theta }{\cos \theta }.2\sin \theta \cos \theta d\theta } $ $ =\int{(1-cos2\theta )d\theta =\theta -\frac{1}{2}\sin 2\theta +c} $ $ ={{\sin }^{-1}}\sqrt{x}-\sqrt{x(1-x)}+c $
BETA
