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受欢迎的微积分 >

integral of sqrt(16-r^2)

解答

\int 16 - r^{2} d r

解答

8 (arcsin (\frac{1}{4} r) + \frac{1}{2} sin (2 arcsin (\frac{1}{4} r))) + C

求解步骤

\int 16 - r^{2} d r

使用三角换元法

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

使用三角恒等式改写

= 16 \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

= 16 \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

= 16 \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)

= 16 \cdot \frac{1}{2} (u + \frac{1}{2} sin (2 u))

u = arcsin (\frac{1}{4} r)

代回

= 16 \cdot \frac{1}{2} (arcsin (\frac{1}{4} r) + \frac{1}{2} sin (2 arcsin (\frac{1}{4} r)))

化简

16 \cdot \frac{1}{2} (arcsin (\frac{1}{4} r) + \frac{1}{2} sin (2 arcsin (\frac{1}{4} r))) : 8 (arcsin (\frac{1}{4} r) + \frac{1}{2} sin (2 arcsin (\frac{1}{4} r)))

= 8 (arcsin (\frac{1}{4} r) + \frac{1}{2} sin (2 arcsin (\frac{1}{4} r)))

解答补常数

= 8 (arcsin (\frac{1}{4} r) + \frac{1}{2} sin (2 arcsin (\frac{1}{4} r))) + C

作图

流行的例子

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\int cos (7 x) cos (4 x) d x

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