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

解答

\int_{0}^{2 π} 4 sin^{2} (t) d t

4 π

十进制

12.56637 \dots

求解步骤

\int_{0}^{2 π} 4 sin^{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}^{2 π} sin^{2} (t) d t

使用三角恒等式改写

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

\int_{0}^{2 π} 1 d t = 2 π

\int_{0}^{2 π} cos (2 t) d t = 0

= 4 \cdot \frac{1}{2} (2 π - 0)

化简

4 \cdot \frac{1}{2} (2 π - 0) : 4 π

= 4 π

\int_{0}^{π} π sin (x) d x

\int_{1}^{3} \frac{ln ^{2} ( 5 )}{2 x} d x

\int_{0}^{1} 5 x 5^{x} d x

\int_{0}^{4} (x + 1) - \frac{x}{2} d x

\int_{0}^{x} t (t - sin (t)) d t