What is the integral of (1-x)sqrt(2x-x^2) ?

The integral of (1-x)sqrt(2x-x^2) is 1/3 (-x^2+2x)^{3/2}+C

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integral of (1-x)sqrt(2x-x^2)

\int (1 - x) 2 x - x^{2} d x

\frac{1}{3} (- x^{2} + 2 x)^{\frac{3}{2}} + C

Solution steps

\int (1 - x) 2 x - x^{2} d x

Expand

(1 - x) 2 x - x^{2} : 2 x - x^{2} - x 2 x - x^{2}

= \int 2 x - x^{2} - x 2 x - x^{2} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int 2 x - x^{2} d x - \int x 2 x - x^{2} d x

\int 2 x - x^{2} d x = \frac{1}{2} (arcsin (x - 1) + \frac{1}{2} sin (2 arcsin (x - 1)))

\int x 2 x - x^{2} d x = - \frac{1}{3} (- x^{2} + 2 x)^{\frac{3}{2}} + \frac{1}{2} (arcsin (x - 1) + \frac{1}{2} sin (2 arcsin (x - 1)))

= \frac{1}{2} (arcsin (x - 1) + \frac{1}{2} sin (2 arcsin (x - 1))) - (- \frac{1}{3} (- x^{2} + 2 x)^{\frac{3}{2}} + \frac{1}{2} (arcsin (x - 1) + \frac{1}{2} sin (2 arcsin (x - 1))))

Simplify

\frac{1}{2} (arcsin (x - 1) + \frac{1}{2} sin (2 arcsin (x - 1))) - (- \frac{1}{3} (- x^{2} + 2 x)^{\frac{3}{2}} + \frac{1}{2} (arcsin (x - 1) + \frac{1}{2} sin (2 arcsin (x - 1)))) : \frac{1}{3} (- x^{2} + 2 x)^{\frac{3}{2}}

= \frac{1}{3} (- x^{2} + 2 x)^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{3} (- x^{2} + 2 x)^{\frac{3}{2}} + C

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What is the integral of (1-x)sqrt(2x-x^2) ?
The integral of (1-x)sqrt(2x-x^2) is 1/3 (-x^2+2x)^{3/2}+C