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) \[{{\sin }^{-1}}\sqrt{x}-\sqrt{x(1-x)}+c\]
Correct Answer: D
Solution :
[d] Put \[x={{\sin }^{2}}\theta \Rightarrow dx=2\sin \theta \cos \theta \] \[\therefore \int{\sqrt{\frac{x}{1-x}dx}}=\int{\frac{\sin \theta }{\cos \theta }.2\sin \theta \cos \theta d\theta }\] \[=\int{(1-cos2\theta )d\theta =\theta -\frac{1}{2}\sin 2\theta +c}\] \[={{\sin }^{-1}}\sqrt{x}-\sqrt{x(1-x)}+c\]
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