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integral of ((3x^2)/(sqrt(64-x^2)))

解答

\int (\frac{3 x ^{2}}{64 - x ^{2}}) d x

解答

96 (arcsin (\frac{1}{8} x) - \frac{1}{2} sin (2 arcsin (\frac{1}{8} x))) + C

求解步骤

\int \frac{3 x ^{2}}{64 - x ^{2}} d x

提出常数:

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

= 3 \cdot \int \frac{x ^{2}}{64 - x ^{2}} d x

使用三角换元法

= 3 \cdot \int 64 sin^{2} (u) d u

提出常数:

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

= 3 \cdot 64 \cdot \int sin^{2} (u) d u

使用三角恒等式改写

= 3 \cdot 64 \cdot \int \frac{1 - cos ( 2 u )}{2} d u

提出常数:

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

= 3 \cdot 64 \cdot \frac{1}{2} \cdot \int 1 - cos (2 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

= 3 \cdot 64 \cdot \frac{1}{2} (\int 1 d u - \int cos (2 u) d u)

\int 1 d u = u

\int cos (2 u) d u = \frac{1}{2} sin (2 u)

= 3 \cdot 64 \cdot \frac{1}{2} (u - \frac{1}{2} sin (2 u))

u = arcsin (\frac{1}{8} x)

代回

= 3 \cdot 64 \cdot \frac{1}{2} (arcsin (\frac{1}{8} x) - \frac{1}{2} sin (2 arcsin (\frac{1}{8} x)))

化简

3 \cdot 64 \cdot \frac{1}{2} (arcsin (\frac{1}{8} x) - \frac{1}{2} sin (2 arcsin (\frac{1}{8} x))) : 96 (arcsin (\frac{1}{8} x) - \frac{1}{2} sin (2 arcsin (\frac{1}{8} x)))

= 96 (arcsin (\frac{1}{8} x) - \frac{1}{2} sin (2 arcsin (\frac{1}{8} x)))

解答补常数

= 96 (arcsin (\frac{1}{8} x) - \frac{1}{2} sin (2 arcsin (\frac{1}{8} x))) + C

作图

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