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/2 of (cos(x))^6

解答

\int_{0}^{\frac{π}{2}} (cos (x))^{6} d x

解答

\frac{5 π}{32}

+1

十进制

0.49087 \dots

求解步骤

\int_{0}^{\frac{π}{2}} (cos (x))^{6} d x

使用换元积分法

= \int_{1}^{0} - \frac{u ^{6}}{1 - u ^{2}} d u

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

= - \int_{0}^{1} - \frac{u ^{6}}{1 - u ^{2}} d u

提出常数:

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

= - (- \int_{0}^{1} \frac{u ^{6}}{1 - u ^{2}} d u)

使用三角换元法

= - (- \int_{0}^{\frac{π}{2}} sin^{6} (v) d v)

使用降次积分法

= - (- (- [\frac{cos ( v ) sin ^{5} ( v )}{6}]_{0}^{\frac{π}{2}} + \frac{5}{6} \cdot \int_{0}^{\frac{π}{2}} sin^{4} (v) d v))

= - (- (- [\frac{cos ( v ) sin ^{5} ( v )}{6}]_{0}^{\frac{π}{2}} + \frac{5}{6} \cdot \int_{0}^{\frac{π}{2}} sin^{4} (v) d v))

使用降次积分法

= - (- (- [\frac{cos ( v ) sin ^{5} ( v )}{6}]_{0}^{\frac{π}{2}} + \frac{5}{6} (- [\frac{cos ( v ) sin ^{3} ( v )}{4}]_{0}^{\frac{π}{2}} + \frac{3}{4} \cdot \int_{0}^{\frac{π}{2}} sin^{2} (v) d v)))

= - (- (- [\frac{cos ( v ) sin ^{5} ( v )}{6}]_{0}^{\frac{π}{2}} + \frac{5}{6} (- [\frac{cos ( v ) sin ^{3} ( v )}{4}]_{0}^{\frac{π}{2}} + \frac{3}{4} \cdot \int_{0}^{\frac{π}{2}} sin^{2} (v) d v)))

使用三角恒等式改写

= - (- (- [\frac{cos ( v ) sin ^{5} ( v )}{6}]_{0}^{\frac{π}{2}} + \frac{5}{6} (- [\frac{cos ( v ) sin ^{3} ( v )}{4}]_{0}^{\frac{π}{2}} + \frac{3}{4} \cdot \int_{0}^{\frac{π}{2}} \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{cos ( v ) sin ^{5} ( v )}{6}]_{0}^{\frac{π}{2}} + \frac{5}{6} (- [\frac{cos ( v ) sin ^{3} ( v )}{4}]_{0}^{\frac{π}{2}} + \frac{3}{4} \cdot \frac{1}{2} \cdot \int_{0}^{\frac{π}{2}} 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{cos ( v ) sin ^{5} ( v )}{6}]_{0}^{\frac{π}{2}} + \frac{5}{6} (- [\frac{cos ( v ) sin ^{3} ( v )}{4}]_{0}^{\frac{π}{2}} + \frac{3}{4} \cdot \frac{1}{2} (\int_{0}^{\frac{π}{2}} 1 d v - \int_{0}^{\frac{π}{2}} cos (2 v) d v))))

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

\int_{0}^{\frac{π}{2}} cos (2 v) d v = 0

= - (- (- [\frac{cos ( v ) sin ^{5} ( v )}{6}]_{0}^{\frac{π}{2}} + \frac{5}{6} (- [\frac{cos ( v ) sin ^{3} ( v )}{4}]_{0}^{\frac{π}{2}} + \frac{3}{4} \cdot \frac{1}{2} (\frac{π}{2} - 0))))

化简

- (- (- [\frac{cos ( v ) sin ^{5} ( v )}{6}]_{0}^{\frac{π}{2}} + \frac{5}{6} (- [\frac{cos ( v ) sin ^{3} ( v )}{4}]_{0}^{\frac{π}{2}} + \frac{3}{4} \cdot \frac{1}{2} (\frac{π}{2} - 0)))) : - [\frac{1}{6} sin^{5} (v) cos (v)]_{0}^{\frac{π}{2}} + \frac{5}{6} (- [\frac{1}{4} sin^{3} (v) cos (v)]_{0}^{\frac{π}{2}} + \frac{3 π}{16})

= - [\frac{1}{6} sin^{5} (v) cos (v)]_{0}^{\frac{π}{2}} + \frac{5}{6} (- [\frac{1}{4} sin^{3} (v) cos (v)]_{0}^{\frac{π}{2}} + \frac{3 π}{16})

带入上下限计算后得:

0

带入上下限计算后得:

0

= - 0 + \frac{5}{6} (- 0 + \frac{3 π}{16})

化简

= \frac{5 π}{32}

作图

流行的例子

integral from-infinity to-27 of x^{-5/3}

\int_{- \infty}^{- 27} x^{- \frac{5}{3}} d x

integral from-1 to 2 of x^3(x^2+1)^8

\int_{- 1}^{2} x^{3} (x^{2} + 1)^{8} d x

integral from-1 to 1 of e^{-jwt}

\int_{- 1}^{1} e^{- j wt} d t

integral from 0 to 1 of (pi)

\int_{0}^{1} (π) d x

integral from 0 to pi of e^{-px}sin(x)

\int_{0}^{π} e^{- p x} sin (x) d x