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受欢迎的微积分 >

integral from 0 to pi/(18) of sin^5(9x)

解答

\int_{0}^{\frac{π}{18}} sin^{5} (9 x) d x

解答

\frac{8}{135}

+1

十进制

0.05925 \dots

求解步骤

\int_{0}^{\frac{π}{18}} sin^{5} (9 x) d x

使用换元积分法

= \int_{0}^{\frac{π}{2}} sin^{5} (u) \frac{1}{9} d u

提出常数:

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

= \frac{1}{9} \cdot \int_{0}^{\frac{π}{2}} sin^{5} (u) d u

使用三角恒等式改写

= \frac{1}{9} \cdot \int_{0}^{\frac{π}{2}} (1 - cos^{2} (u))^{2} sin (u) d u

使用换元积分法

= \frac{1}{9} \cdot \int_{1}^{0} - 1 + 2 v^{2} - v^{4} d v

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= \frac{1}{9} (- \int_{0}^{1} - 1 + 2 v^{2} - v^{4} 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{1}{9} (- (- \int_{0}^{1} 1 d v + \int_{0}^{1} 2 v^{2} d v - \int_{0}^{1} v^{4} d v))

\int_{0}^{1} 1 d v = 1

\int_{0}^{1} 2 v^{2} d v = \frac{2}{3}

\int_{0}^{1} v^{4} d v = \frac{1}{5}

= \frac{1}{9} (- (- 1 + \frac{2}{3} - \frac{1}{5}))

化简

\frac{1}{9} (- (- 1 + \frac{2}{3} - \frac{1}{5})) : \frac{8}{135}

= \frac{8}{135}

作图

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