A) \[{{\tan }^{-1}}(2x)+c\]
B) \[{{\cot }^{-1}}(2x)+c\]
C) \[{{\cos }^{-1}}(2x)+c\]
D) \[{{\sin }^{-1}}(2x)+c\]
Correct Answer: D
Solution :
\[I=\int{\frac{2dx}{\sqrt{1-4{{x}^{2}}}}}\]. 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, \[I={{\sin }^{-1}}(2x)+c.\]
You need to login to perform this action.
You will be redirected in
3 sec
