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受欢迎的微积分 >

integral from 0 to pi/6 of sqrt(1-cos(6x))

解答

\int_{0}^{\frac{π}{6}} 1 - cos (6 x) d x

解答

\frac{2}{3}

+1

十进制

0.47140 \dots

求解步骤

\int_{0}^{\frac{π}{6}} 1 - cos (6 x) d x

使用换元积分法

= \int_{0}^{\frac{π}{3}} \frac{1}{2} 1 - cos (3 u) d u

提出常数:

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

= \frac{1}{2} \cdot \int_{0}^{\frac{π}{3}} 1 - cos (3 u) d u

使用换元积分法

= \frac{1}{2} \cdot \int_{0}^{\frac{2 π}{3}} \frac{1}{2} 1 - cos (\frac{3 v}{2}) d v

提出常数:

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

= \frac{1}{2} \cdot \frac{1}{2} \cdot \int_{0}^{\frac{2 π}{3}} 1 - cos (\frac{3 v}{2}) d v

使用换元积分法

= \frac{1}{2} \cdot \frac{1}{2} \cdot \int_{0}^{π} 1 - cos (w) \frac{2}{3} d w

提出常数:

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

= \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{2}{3} \cdot \int_{0}^{π} 1 - cos (w) d w

使用三角恒等式改写

= \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{2}{3} \cdot \int_{0}^{π} 2 sin^{2} (\frac{w}{2}) d w

提出常数:

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

= \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{2}{3} 2 \cdot \int_{0}^{π} sin^{2} (\frac{w}{2}) d w

使用指数法则:

a^{2} = ∣ a ∣

= \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{2}{3} 2 \cdot \int_{0}^{π} sin (\frac{w}{2}) d w

拿掉绝对值

= \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{2}{3} 2 \cdot \int_{0}^{π} sin (\frac{w}{2}) d w

使用换元积分法

= \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{2}{3} 2 \cdot \int_{0}^{\frac{π}{2}} sin (t) \cdot 2 d t

提出常数:

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

= \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{2}{3} 2 \cdot 2 \cdot \int_{0}^{\frac{π}{2}} sin (t) d t

使用积分常见定则:

\int sin (t) d t = - cos (t)

= \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{2}{3} 2 \cdot 2 [- cos (t)]_{0}^{\frac{π}{2}}

化简

\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{2}{3} 2 \cdot 2 [- cos (t)]_{0}^{\frac{π}{2}} : \frac{2}{3} [- cos (t)]_{0}^{\frac{π}{2}}

= \frac{2}{3} [- cos (t)]_{0}^{\frac{π}{2}}

带入上下限计算后得:

1

= \frac{2}{3} \cdot 1

化简

= \frac{2}{3}

作图

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