Question

\[\int\limits_0^4 \frac{1}{\sqrt{4x - x^2}} dx\]

Answer 1

\[Let\ I = \int_0^4 \frac{1}{\sqrt{4x - x^2}} d x . Then, \]
\[I = \int_0^4 \frac{1}{\sqrt{4x - x^2 - 4 + 4}} d x\]
\[ \Rightarrow I = \int_0^4 \frac{1}{\sqrt{- \left( x - 2 \right)^2 + 4}} d x\]
\[ \Rightarrow I = \left[ \sin^{- 1} \frac{\left( x - 2 \right)}{2} \right]_0^4 \]
\[ \Rightarrow I = \left( \sin^{- 1} 1 - \sin^{- 1} ( - 1) \right)\]
\[ \Rightarrow I = 2 \sin^{- 1} 1\]
\[ \Rightarrow I = 2 \frac{\pi}{2} = \pi\]