integral from 0 to pi of 4cos^2(t)

解答

\int_{0}^{π} 4 cos^{2} (t) d t

2 π

十进制

6.28318 \dots

求解步骤

\int_{0}^{π} 4 cos^{2} (t) d t

提出常数:

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

= 4 \cdot \int_{0}^{π} cos^{2} (t) d t

使用三角恒等式改写

= 4 \cdot \int_{0}^{π} \frac{1 + cos ( 2 t )}{2} d t

提出常数:

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

= 4 \cdot \frac{1}{2} \cdot \int_{0}^{π} 1 + cos (2 t) d t

使用积分加法定则:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 4 \cdot \frac{1}{2} (\int_{0}^{π} 1 d t + \int_{0}^{π} cos (2 t) d t)

\int_{0}^{π} 1 d t = π

\int_{0}^{π} cos (2 t) d t = 0

= 4 \cdot \frac{1}{2} (π + 0)

化简

4 \cdot \frac{1}{2} (π + 0) : 2 π

= 2 π