A) \[2\sqrt{1-x}+{{\cos }^{-1}}\sqrt{x}+\sqrt{x-{{x}^{2}}}+c\]
B) \[-2\sqrt{1-x}+{{\cos }^{-1}}\sqrt{x}+\sqrt{x-{{x}^{2}}}+c\]
C) \[2\sqrt{1-x}-{{\cos }^{-1}}\sqrt{x}-\sqrt{x-{{x}^{2}}}+c\]
D) None of these
Correct Answer: B
Solution :
[b] \[I=\int{\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}.dx}\] Let \[x=co{{s}^{2}}\theta \] \[\therefore dx=-2sin\theta .cos\theta .d\theta \] Now, \[I=\int{\sqrt{\frac{1-\cos \theta }{1+\cos \theta }}}.(-2\sin \theta .\cos \theta ).d\theta \] \[=-2\int{\frac{\sin \frac{\theta }{2}}{\cos \frac{\theta }{2}}.\left( 2.\sin \frac{\theta }{2}.\cos \frac{\theta }{2}.\cos \theta \right)}d\theta \] \[=-2\int{2{{\sin }^{2}}\frac{\theta }{2}.\cos \theta .d\theta =-2\int{(1-\cos \theta ).\cos \theta .d\theta }}\] \[=-2\int{\cos \theta d\theta +2}\int{{{\cos }^{2}}\theta d\theta }\] \[=-2\sin \theta \int{\left( \frac{1+\cos 2\theta }{2} \right)}.d\theta =-2\sin \theta +\frac{1}{2}.\theta +\frac{1}{2}.\frac{\sin 2\theta }{2}+c\]\[=-2.\sqrt{1-x}+{{\cos }^{-1}}\sqrt{x}+\sin \theta .\cos \theta +c\] \[=-2\sqrt{1-x}+{{\cos }^{-1}}\sqrt{x}+\sqrt{1-x}.\sqrt{x}+c\] \[=-2.\sqrt{1-x}+{{\cos }^{-1}}\sqrt{x}+\sqrt{x-{{x}^{2}}}+c\] Hence, option [b] is correct.
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