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प्रश्न
\[\int\sqrt{2x - x^2} \text{ dx}\]
योग
उत्तर
\[I = \int\sqrt{2x - x^2}\text{ dx}\]
\[ = \int\sqrt{x\left( 2 - x \right)}\text{ dx}\]
\[ = \int\sqrt{x\left( 2 - x \right)}\text{ dx}\]
Let
\[x = 1 + \ sin\ u\]
\[or, dx = \cos\ u\ du\]
\[ \Rightarrow I = \int\sqrt{\left( 1 + \sin u \right)\left( 1 - \sin u \right)}\ cos\ u\ du\]
\[ \Rightarrow I = \int \cos^2 u\ du\]
\[ \Rightarrow I = \frac{1}{2}\int\left( \cos2u + 1 \right)du\]
\[ \Rightarrow I = \int\sqrt{\left( 1 + \sin u \right)\left( 1 - \sin u \right)}\ cos\ u\ du\]
\[ \Rightarrow I = \int \cos^2 u\ du\]
\[ \Rightarrow I = \frac{1}{2}\int\left( \cos2u + 1 \right)du\]
\[\Rightarrow I = \frac{1}{2}\left( \frac{1}{2}\sin 2u + u \right) + c\]
\[ \Rightarrow I = \frac{1}{2}\left( \sin u \cos u + u \right) + c\]
\[ \Rightarrow I = \frac{1}{2}\left( \sin u \sqrt{1 - \sin^2 u} + u \right) + c\]
\[ \therefore I = \frac{1}{2}\left( x - 1 \right)\sqrt{2x - x^2} + \frac{1}{2} \sin^{- 1} \left( x - 1 \right) + c \left[ \because u = \sin^{- 1} \left( x - 1 \right) \right]\]
\[ \Rightarrow I = \frac{1}{2}\left( \sin u \cos u + u \right) + c\]
\[ \Rightarrow I = \frac{1}{2}\left( \sin u \sqrt{1 - \sin^2 u} + u \right) + c\]
\[ \therefore I = \frac{1}{2}\left( x - 1 \right)\sqrt{2x - x^2} + \frac{1}{2} \sin^{- 1} \left( x - 1 \right) + c \left[ \because u = \sin^{- 1} \left( x - 1 \right) \right]\]
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