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受欢迎的微积分 >

integral of sqrt(6-4x-4x^2)

解答

\int 6 - 4 x - 4 x^{2} d x

解答

\frac{7}{4} (arcsin (\frac{1}{7} (2 x + 1)) + \frac{1}{2} sin (2 arcsin (\frac{2 x + 1}{7}))) + C

求解步骤

\int 6 - 4 x - 4 x^{2} d x

对

6 - 4 x - 4 x^{2}

配方:

- 4 (x + \frac{1}{2})^{2} + 7

= \int - 4 (x + \frac{1}{2})^{2} + 7 d x

使用换元积分法

= \int - 4 u^{2} + 7 d u

使用三角换元法

= \int \frac{7}{2} cos^{2} (v) d v

提出常数:

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

= \frac{7}{2} \cdot \int cos^{2} (v) d v

使用三角恒等式改写

= \frac{7}{2} \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

= \frac{7}{2} \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

= \frac{7}{2} \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)

= \frac{7}{2} \cdot \frac{1}{2} (v + \frac{1}{2} sin (2 v))

代回

= \frac{7}{2} \cdot \frac{1}{2} (arcsin (\frac{2}{7} (x + \frac{1}{2})) + \frac{1}{2} sin (2 arcsin (\frac{2}{7} (x + \frac{1}{2}))))

化简

\frac{7}{2} \cdot \frac{1}{2} (arcsin (\frac{2}{7} (x + \frac{1}{2})) + \frac{1}{2} sin (2 arcsin (\frac{2}{7} (x + \frac{1}{2})))) : \frac{7}{4} (arcsin (\frac{1}{7} (2 x + 1)) + \frac{1}{2} sin (2 arcsin (\frac{2 x + 1}{7})))

= \frac{7}{4} (arcsin (\frac{1}{7} (2 x + 1)) + \frac{1}{2} sin (2 arcsin (\frac{2 x + 1}{7})))

解答补常数

= \frac{7}{4} (arcsin (\frac{1}{7} (2 x + 1)) + \frac{1}{2} sin (2 arcsin (\frac{2 x + 1}{7}))) + C

作图

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