integral from 0 to pi/6 of 64sin^2(x)

解答

\int_{0}^{\frac{π}{6}} 64 sin^{2} (x) d x

32 (\frac{π}{6} - \frac{3}{4})

十进制

2.89875 \dots

求解步骤

\int_{0}^{\frac{π}{6}} 64 sin^{2} (x) d x

提出常数:

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

= 64 \cdot \int_{0}^{\frac{π}{6}} sin^{2} (x) d x

使用三角恒等式改写

= 64 \cdot \int_{0}^{\frac{π}{6}} \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

= 64 \cdot \frac{1}{2} \cdot \int_{0}^{\frac{π}{6}} 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

= 64 \cdot \frac{1}{2} (\int_{0}^{\frac{π}{6}} 1 d x - \int_{0}^{\frac{π}{6}} cos (2 x) d x)

\int_{0}^{\frac{π}{6}} 1 d x = \frac{π}{6}

\int_{0}^{\frac{π}{6}} cos (2 x) d x = \frac{3}{4}

= 64 \cdot \frac{1}{2} (\frac{π}{6} - \frac{3}{4})

化简

64 \cdot \frac{1}{2} (\frac{π}{6} - \frac{3}{4}) : 32 (\frac{π}{6} - \frac{3}{4})

= 32 (\frac{π}{6} - \frac{3}{4})

\int \frac{cos ( \frac{π}{x ^{31}} )}{x ^{32}} d x

d er i v a t i v e y = \frac{e ^{x}}{1 + e ^{x}}

\frac{d}{d x} (\frac{x ^{2} + 2 x - 23}{x - 4})

d er i v a t i v e y = 9 cos (x)

d er i v a t i v e \frac{x}{x + 2}