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integral of sqrt(5+4x-x^2)

解答

\int 5 + 4 x - x^{2} d x

解答

\frac{9}{2} (arcsin (\frac{1}{3} (x - 2)) + \frac{1}{2} sin (2 arcsin (\frac{1}{3} (x - 2)))) + C

求解步骤

\int 5 + 4 x - x^{2} d x

对

5 + 4 x - x^{2}

配方:

- (x - 2)^{2} + 9

= \int - (x - 2)^{2} + 9 d x

使用换元积分法

= \int - u^{2} + 9 d u

使用三角换元法

= \int 9 cos^{2} (v) d v

提出常数:

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

= 9 \cdot \int cos^{2} (v) d v

使用三角恒等式改写

= 9 \cdot \int \frac{1 + cos ( 2 v )}{2} d v

提出常数:

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

= 9 \cdot \frac{1}{2} \cdot \int 1 + cos (2 v) d v

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

= 9 \cdot \frac{1}{2} (\int 1 d v + \int cos (2 v) d v)

\int 1 d v = v

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

= 9 \cdot \frac{1}{2} (v + \frac{1}{2} sin (2 v))

代回

= 9 \cdot \frac{1}{2} (arcsin (\frac{1}{3} (x - 2)) + \frac{1}{2} sin (2 arcsin (\frac{1}{3} (x - 2))))

化简

9 \cdot \frac{1}{2} (arcsin (\frac{1}{3} (x - 2)) + \frac{1}{2} sin (2 arcsin (\frac{1}{3} (x - 2)))) : \frac{9}{2} (arcsin (\frac{1}{3} (x - 2)) + \frac{1}{2} sin (2 arcsin (\frac{1}{3} (x - 2))))

= \frac{9}{2} (arcsin (\frac{1}{3} (x - 2)) + \frac{1}{2} sin (2 arcsin (\frac{1}{3} (x - 2))))

解答补常数

= \frac{9}{2} (arcsin (\frac{1}{3} (x - 2)) + \frac{1}{2} sin (2 arcsin (\frac{1}{3} (x - 2)))) + C

作图

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