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受欢迎的微积分 >

integral from 0 to 2pi of 2cos(x)sin(x)

解答

\int_{0}^{2 π} 2 cos (x) sin (x) d x

解答

0

求解步骤

\int_{0}^{2 π} 2 cos (x) sin (x) d x

提出常数:

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

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

使用三角恒等式改写

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

提出常数:

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

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

使用换元积分法

= 2 \cdot \frac{1}{2} \cdot \int_{0}^{4 π} sin (u) \frac{1}{2} d u

提出常数:

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

= 2 \cdot \frac{1}{2} \cdot \frac{1}{2} \cdot \int_{0}^{4 π} sin (u) d u

使用积分常见定则:

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

= 2 \cdot \frac{1}{2} \cdot \frac{1}{2} [- cos (u)]_{0}^{4 π}

化简

2 \cdot \frac{1}{2} \cdot \frac{1}{2} [- cos (u)]_{0}^{4 π} : \frac{1}{2} [- cos (u)]_{0}^{4 π}

= \frac{1}{2} [- cos (u)]_{0}^{4 π}

带入上下限计算后得:

0

= \frac{1}{2} \cdot 0

化简

= 0

作图

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