integral from 0 to pi of 16sin^4(x)

解答

\int_{0}^{π} 16 sin^{4} (x) d x

6 π

十进制

18.84955 \dots

求解步骤

\int_{0}^{π} 16 sin^{4} (x) d x

提出常数:

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

= 16 \cdot \int_{0}^{π} sin^{4} (x) d x

使用降次积分法

= 16 (- [\frac{cos ( x ) sin ^{3} ( x )}{4}]_{0}^{π} + \frac{3}{4} \cdot \int_{0}^{π} sin^{2} (x) d x)

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

= 16 (- [\frac{cos ( x ) sin ^{3} ( x )}{4}]_{0}^{π} + \frac{3}{4} \cdot \frac{π}{2})

化简

16 (- [\frac{cos ( x ) sin ^{3} ( x )}{4}]_{0}^{π} + \frac{3}{4} \cdot \frac{π}{2}) : 16 (- [\frac{1}{4} sin^{3} (x) cos (x)]_{0}^{π} + \frac{3 π}{8})

= 16 (- [\frac{1}{4} sin^{3} (x) cos (x)]_{0}^{π} + \frac{3 π}{8})

带入上下限计算后得:

0

= 16 (- 0 + \frac{3 π}{8})

化简

= 6 π

x^{3} y^{^{'}} + 2 x^{2} y = - 2 cos (2 x) y^{- \frac{1}{2}}

d er i v a t i v e f (x) = (x^{2} + 3 x) (x - 1)

\frac{d}{d x} (120 x^{3} - 180 x^{2} + 60)

in v erse l a pl a ce \frac{6 e ^{- s}}{s ( 10 s + 1 )}

\frac{\partial}{\partial x} (\frac{( 3 )}{( x + y ) ^{3}})