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受欢迎的微积分 >

integral from 0 to 2pi of 3sin(t)cos(t)

解答

\int_{0}^{2 π} 3 sin (t) cos (t) d t

解答

0

求解步骤

\int_{0}^{2 π} 3 sin (t) cos (t) d t

提出常数:

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

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

使用三角恒等式改写

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

提出常数:

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

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

使用换元积分法

= 3 \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

= 3 \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)

= 3 \cdot \frac{1}{2} \cdot \frac{1}{2} [- cos (u)]_{0}^{4 π}

化简

3 \cdot \frac{1}{2} \cdot \frac{1}{2} [- cos (u)]_{0}^{4 π} : \frac{3}{4} [- cos (u)]_{0}^{4 π}

= \frac{3}{4} [- cos (u)]_{0}^{4 π}

带入上下限计算后得:

0

= \frac{3}{4} \cdot 0

化简

= 0

作图

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