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受欢迎的微积分 >

integral from 0 to 2pi of sqrt((1-cos(x))/2)

解答

\int_{0}^{2 π} \frac{1 - cos ( x )}{2} d x

解答

4

求解步骤

\int_{0}^{2 π} \frac{1 - cos ( x )}{2} d x

\frac{1 - cos ( x )}{2} = \frac{1}{2} 1 - cos (x)

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

提出常数:

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

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

使用三角恒等式改写

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

提出常数:

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

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

使用指数法则:

a^{2} = ∣ a ∣

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

拿掉绝对值

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

使用换元积分法

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

提出常数:

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

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

使用积分常见定则:

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

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

化简

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

= 2 [- cos (u)]_{0}^{π}

带入上下限计算后得:

2

= 2 \cdot 2

化简

= 4

作图

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