A) \[\pi +2\]
B) \[\pi +\frac{3}{2}\]
C) \[\pi +1\]
D) None of these
Correct Answer: A
Solution :
Put \[x=2\cos \theta \Rightarrow dx=-2\sin \theta \,d\theta ,\]then \[\int_{0}^{2}{\sqrt{\frac{2+x}{2-x}}}dx=-2\int_{\pi /2}^{0}{\sqrt{\frac{1+\cos \theta }{1-\cos \theta }}}\sin \theta \,d\theta \] \[=4\int_{0}^{\pi /2}{\frac{\cos (\theta /2)}{\sin (\theta /2)}\sin \frac{\theta }{2}\cos \frac{\theta }{2}d\theta }\] \[=2\int_{0}^{\pi /2}{(1+\cos \theta )\,d\theta }\] \[=2[\theta +\sin \theta ]_{0}^{\pi /2}=2\left[ \frac{\pi }{2}+1 \right]=\pi +2\].
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