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Question

Question: \[\int_{}^{}\sqrt{2ax - x^{2}}dx =\]...

∫2ax−x2dx=\int_{}^{}\sqrt{2ax - x^{2}}dx =

A

12(x−a)2ax−x2+12a2sin⁡−1(x−aa)+c\frac{1}{2}(x - a)\sqrt{2ax - x^{2}} + \frac{1}{2}a^{2}\sin^{- 1}\left( \frac{x - a}{a} \right) + c

B

12(x−a)2ax−x2−12a2sin⁡−1(x−aa)+c\frac{1}{2}(x - a)\sqrt{2ax - x^{2}} - \frac{1}{2}a^{2}\sin^{- 1}\left( \frac{x - a}{a} \right) + c

C

12(x−a)2ax−x2+12a2cos⁡−1(x−aa)+c\frac{1}{2}(x - a)\sqrt{2ax - x^{2}} + \frac{1}{2}a^{2}\cos^{- 1}\left( \frac{x - a}{a} \right) + c

D

12(x−a)2ax−x2−12a2cos⁡−1(x−aa)+c\frac{1}{2}(x - a)\sqrt{2ax - x^{2}} - \frac{1}{2}a^{2}\cos^{- 1}\left( \frac{x - a}{a} \right) + c

Answer

12(x−a)2ax−x2+12a2sin⁡−1(x−aa)+c\frac{1}{2}(x - a)\sqrt{2ax - x^{2}} + \frac{1}{2}a^{2}\sin^{- 1}\left( \frac{x - a}{a} \right) + c

Explanation

Solution

∫2ax−x2dx\int_{}^{}\sqrt{2ax - x^{2}}dx

=∫a2−a2+2ax−x2dx=∫a2−(x2−2ax+a2)dx= \int_{}^{}{\sqrt{a^{2} - a^{2} + 2ax - x^{2}}dx}\mathbf{=}\int_{}^{}\sqrt{\mathbf{a}^{\mathbf{2}}\mathbf{- (}\mathbf{x}^{\mathbf{2}}\mathbf{- 2ax +}\mathbf{a}^{\mathbf{2}}\mathbf{)}}\mathbf{dx} =∫a2−(x−a)2dx\mathbf{=}\int_{}^{}{\sqrt{\mathbf{a}^{\mathbf{2}}\mathbf{-}\mathbf{(x}\mathbf{-}\mathbf{a}\mathbf{)}^{\mathbf{2}}}\mathbf{dx}}

=12(x−a)2ax−x2+12a2sin⁡−1(x−a)a+c\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{(x}\mathbf{-}\mathbf{a)}\sqrt{\mathbf{2ax}\mathbf{-}\mathbf{x}^{\mathbf{2}}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{a}^{\mathbf{2}}\mathbf{\sin}^{\mathbf{-}\mathbf{1}}\frac{\mathbf{(x}\mathbf{-}\mathbf{a)}}{\mathbf{a}}\mathbf{+ c}