zs

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

受欢迎的微积分 >

integral from 0 to pi/(10) of sin^5(5x)

解答

\int_{0}^{\frac{π}{10}} sin^{5} (5 x) d x

解答

\frac{8}{75}

+1

十进制

0.10666 \dots

求解步骤

\int_{0}^{\frac{π}{10}} sin^{5} (5 x) d x

使用换元积分法

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

提出常数:

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

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

使用三角恒等式改写

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

使用换元积分法

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

化简

\frac{1}{5} (- (- 1 + \frac{2}{3} - \frac{1}{5})) : \frac{8}{75}

= \frac{8}{75}

作图

流行的例子

integral from-pi to pi of (x+pi)

\int_{- π}^{π} (x + π) d x

integral from-1 to 1 of 1/(x^{5/3)}

\int_{- 1}^{1} \frac{1}{x ^{\frac{5}{3}}} d x

integral from 0 to x of sqrt(x^2+y^2)

\int_{0}^{x} x^{2} + y^{2} d y

integral from-7 to 1 of x+3

\int_{- 7}^{1} x + 3 d x

integral from 1/3 to 2/3 of sqrt(4-9x^2)

\int_{\frac{1}{3}}^{\frac{2}{3}} 4 - 9 x^{2} d x