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受欢迎的微积分 >

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

解答

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

解答

\frac{5}{8} arcsin (x - 1) + \frac{1}{4} sin (2 arcsin (x - 1)) - \frac{1}{32} sin (4 arcsin (x - 1)) - \frac{2}{3} (- x^{2} + 2 x)^{\frac{3}{2}} + C

求解步骤

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

对

2 x - x^{2}

配方:

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

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

使用换元积分法

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

乘开

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

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

使用积分加法定则:

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

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

\int u^{2} - u^{2} + 1 d u = \frac{1}{8} (arcsin (u) - \frac{1}{4} sin (4 arcsin (u)))

\int 2 u - u^{2} + 1 d u = - \frac{2}{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}{8} (arcsin (u) - \frac{1}{4} sin (4 arcsin (u))) - \frac{2}{3} (- u^{2} + 1)^{\frac{3}{2}} + \frac{1}{2} (arcsin (u) + \frac{1}{2} sin (2 arcsin (u)))

u = x - 1

代回

= \frac{1}{8} (arcsin (x - 1) - \frac{1}{4} sin (4 arcsin (x - 1))) - \frac{2}{3} (- (x - 1)^{2} + 1)^{\frac{3}{2}} + \frac{1}{2} (arcsin (x - 1) + \frac{1}{2} sin (2 arcsin (x - 1)))

化简

\frac{1}{8} (arcsin (x - 1) - \frac{1}{4} sin (4 arcsin (x - 1))) - \frac{2}{3} (- (x - 1)^{2} + 1)^{\frac{3}{2}} + \frac{1}{2} (arcsin (x - 1) + \frac{1}{2} sin (2 arcsin (x - 1))) : \frac{5}{8} arcsin (x - 1) + \frac{1}{4} sin (2 arcsin (x - 1)) - \frac{1}{32} sin (4 arcsin (x - 1)) - \frac{2}{3} (- x^{2} + 2 x)^{\frac{3}{2}}

= \frac{5}{8} arcsin (x - 1) + \frac{1}{4} sin (2 arcsin (x - 1)) - \frac{1}{32} sin (4 arcsin (x - 1)) - \frac{2}{3} (- x^{2} + 2 x)^{\frac{3}{2}}

解答补常数

= \frac{5}{8} arcsin (x - 1) + \frac{1}{4} sin (2 arcsin (x - 1)) - \frac{1}{32} sin (4 arcsin (x - 1)) - \frac{2}{3} (- x^{2} + 2 x)^{\frac{3}{2}} + C

作图

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