integral from 0 to 2pi of 2a(sin(t/2))

解答

\int_{0}^{2 π} 2 a (sin (\frac{t}{2})) d t

8 a

求解步骤

\int_{0}^{2 π} 2 a sin (\frac{t}{2}) d t

提出常数:

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

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

使用换元积分法

= 2 a \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

= 2 a \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)

= 2 a \cdot 2 [- cos (u)]_{0}^{π}

化简

= 4 a [- cos (u)]_{0}^{π}

带入上下限计算后得:

2

= 4 a \cdot 2

化简

= 8 a