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Question
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Solution
\[Let\ I = \int_0^2 \frac{1}{\sqrt{3 + 2x - x^2}} d x . Then, \]
\[I = \int_0^2 \frac{1}{\sqrt{- x^2 + 2x - 1 + 1 + 3}} d x\]
\[ \Rightarrow I = \int_0^2 \frac{1}{\sqrt{- \left( x - 1 \right)^2 + 4}} d x\]
\[ \Rightarrow I = \left[ \sin^{- 1} \frac{\left( x - 1 \right)}{2} \right]_0^2 \]
\[ \Rightarrow I = \sin^{- 1} \frac{1}{2} - \sin^{- 1} \left( - \frac{1}{2} \right)\]
\[ \Rightarrow I = 2 \sin^{- 1} \frac{1}{2}\]
\[ \Rightarrow I = 2 \times \frac{\pi}{6} = \frac{\pi}{3}\]
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