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

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

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integral of xsqrt(2x-x^2)

\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))) + C

Solution steps

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

Complete the square

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

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

Apply u-substitution

: \int (u + 1) - u^{2} + 1 d u

= \int (u + 1) - u^{2} + 1 d u

Expand

(u + 1) - u^{2} + 1 : u - u^{2} + 1 + - u^{2} + 1

= \int u - u^{2} + 1 + - u^{2} + 1 d u

Apply the Sum Rule:

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

= \int u - u^{2} + 1 d u + \int - u^{2} + 1 d u

\int u - u^{2} + 1 d u = - \frac{1}{3} (- u^{2} + 1)^{\frac{3}{2}}

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

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

Substitute back

u = x - 1

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

Simplify

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

Add a constant to the solution

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

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sin (π x), \frac{1}{2}

(cos^{2} (t))^{^{'}}

\frac{d}{d x} (9 x^{7} + 3 x^{5} - 6 x^{6} - 2 x^{4})

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