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受欢迎的微积分 >

integral of x^2*sqrt(4-x^2)

解答

\int x^{2} \cdot 4 - x^{2} d x

解答

2 (arcsin (\frac{1}{2} x) - \frac{1}{4} sin (4 arcsin (\frac{1}{2} x))) + C

求解步骤

\int x^{2} 4 - x^{2} d x

使用三角换元法

= \int 16 sin^{2} (u) cos^{2} (u) d u

提出常数:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 16 \cdot \int sin^{2} (u) cos^{2} (u) d u

使用三角恒等式改写

= 16 \cdot \int \frac{1 - cos ( 4 u )}{8} d u

提出常数:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 16 \cdot \frac{1}{8} \cdot \int 1 - cos (4 u) 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

= 16 \cdot \frac{1}{8} (\int 1 d u - \int cos (4 u) d u)

\int 1 d u = u

\int cos (4 u) d u = \frac{1}{4} sin (4 u)

= 16 \cdot \frac{1}{8} (u - \frac{1}{4} sin (4 u))

u = arcsin (\frac{1}{2} x)

代回

= 16 \cdot \frac{1}{8} (arcsin (\frac{1}{2} x) - \frac{1}{4} sin (4 arcsin (\frac{1}{2} x)))

化简

16 \cdot \frac{1}{8} (arcsin (\frac{1}{2} x) - \frac{1}{4} sin (4 arcsin (\frac{1}{2} x))) : 2 (arcsin (\frac{1}{2} x) - \frac{1}{4} sin (4 arcsin (\frac{1}{2} x)))

= 2 (arcsin (\frac{1}{2} x) - \frac{1}{4} sin (4 arcsin (\frac{1}{2} x)))

解答补常数

= 2 (arcsin (\frac{1}{2} x) - \frac{1}{4} sin (4 arcsin (\frac{1}{2} x))) + C

作图

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