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

解答

\int_{0}^{2 π} sin^{2} (x) d x

π

十进制

3.14159 \dots

求解步骤

\int_{0}^{2 π} sin^{2} (x) d x

使用三角恒等式改写

= \int_{0}^{2 π} \frac{1 - cos ( 2 x )}{2} d x

提出常数:

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

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

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

= \frac{1}{2} (\int_{0}^{2 π} 1 d x - \int_{0}^{2 π} cos (2 x) d x)

\int_{0}^{2 π} 1 d x = 2 π

\int_{0}^{2 π} cos (2 x) d x = 0

= \frac{1}{2} (2 π - 0)

化简

\frac{1}{2} (2 π - 0) : π

= π

n = 1 \sum \infty n^{- \frac{1}{5}}

x \to 2 - lim (\frac{5 x}{x - 6})

d er i v a t i v e f (x) = (x^{3} + 2) e^{x}

x \to - 2 lim (x^{2} - 4 x + 6)

d er i v a t i v e ln (2 x - 5)