Integral Calculus Question 478
Question: The value of $ \int{\frac{2dx}{\sqrt{1-4x^{2}}}} $ is
[Karnataka CET 2001; Pb. CET 2001]
Options:
A) $ {{\tan }^{-1}}(2x)+c $
B) $ {{\cot }^{-1}}(2x)+c $
C) $ {{\cos }^{-1}}(2x)+c $
D) $ {{\sin }^{-1}}(2x)+c $
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Answer:
Correct Answer: D
Solution:
$ I=\int{\frac{2dx}{\sqrt{1-4x^{2}}}} $ . Put $ 2x=\sin \theta $
Þ $ 2dx=\cos \theta d\theta $
$ \Rightarrow I=\int{\frac{\cos \theta d\theta }{\sqrt{1-{{\sin }^{2}}\theta }}=\int{\frac{\cos \theta }{\cos \theta }d\theta =\int{d\theta +c=\theta +c}}} $ . Therefore, $ I={{\sin }^{-1}}(2x)+c. $
BETA
