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integral of 2sqrt(4-y^2)

解答

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

解答

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

求解步骤

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

提出常数:

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

= 2 \cdot \int 4 - y^{2} d y

使用三角换元法

= 2 \cdot \int 4 cos^{2} (u) d u

提出常数:

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

= 2 \cdot 4 \cdot \int cos^{2} (u) d u

使用三角恒等式改写

= 2 \cdot 4 \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

= 2 \cdot 4 \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

= 2 \cdot 4 \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)

= 2 \cdot 4 \cdot \frac{1}{2} (u + \frac{1}{2} sin (2 u))

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

代回

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

化简

2 \cdot 4 \cdot \frac{1}{2} (arcsin (\frac{1}{2} y) + \frac{1}{2} sin (2 arcsin (\frac{1}{2} y))) : 4 (arcsin (\frac{1}{2} y) + \frac{1}{2} sin (2 arcsin (\frac{1}{2} y)))

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

解答补常数

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

作图

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