integral from 0 to pi of 5sin^2(xco)s^4x

解答

\int_{0}^{π} 5 sin^{2} (x co) s^{4} x d x

\frac{5 s ^{4} ( 2 π ^{2} c ^{2} o ^{2} - 2 π co sin ( 2 π co ) - cos ( 2 π co ) + 1 )}{8 c ^{2} o ^{2}}

求解步骤

\int_{0}^{π} 5 sin^{2} (x co) s^{4} x d x

提出常数:

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

= 5 s^{4} \cdot \int_{0}^{π} sin^{2} (co x) x d x

使用换元积分法

= 5 s^{4} \cdot \int_{0}^{co π} \frac{u sin ^{2} ( u )}{c ^{2} o ^{2}} d u

提出常数:

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

= 5 s^{4} \frac{1}{c ^{2} o ^{2}} \cdot \int_{0}^{co π} u sin^{2} (u) d u

使用分布积分法

= 5 s^{4} \frac{1}{c ^{2} o ^{2}} [\frac{1}{2} u (u - \frac{1}{2} sin (2 u)) - \int \frac{1}{2} (u - \frac{1}{2} sin (2 u)) d u]_{0}^{π co}

\int \frac{1}{2} (u - \frac{1}{2} sin (2 u)) d u = \frac{1}{2} (\frac{u ^{2}}{2} + \frac{1}{4} cos (2 u))

= 5 s^{4} \frac{1}{c ^{2} o ^{2}} [\frac{1}{2} u (u - \frac{1}{2} sin (2 u)) - \frac{1}{2} (\frac{u ^{2}}{2} + \frac{1}{4} cos (2 u))]_{0}^{π co}

化简

5 s^{4} \frac{1}{c ^{2} o ^{2}} [\frac{1}{2} u (u - \frac{1}{2} sin (2 u)) - \frac{1}{2} (\frac{u ^{2}}{2} + \frac{1}{4} cos (2 u))]_{0}^{π co} : \frac{5 s ^{4}}{c ^{2} o ^{2}} [\frac{2 u ^{2} - 2 u sin ( 2 u ) - cos ( 2 u )}{8}]_{0}^{π co}

= \frac{5 s ^{4}}{c ^{2} o ^{2}} [\frac{2 u ^{2} - 2 u sin ( 2 u ) - cos ( 2 u )}{8}]_{0}^{π co}

带入上下限计算后得:

\frac{2 π ^{2} c ^{2} o ^{2} - 2 π co sin ( 2 π co ) - cos ( 2 π co )}{8} + \frac{1}{8}

= \frac{5 s ^{4}}{c ^{2} o ^{2}} (\frac{2 π ^{2} c ^{2} o ^{2} - 2 π co sin ( 2 π co ) - cos ( 2 π co )}{8} + \frac{1}{8})

化简

= \frac{5 s ^{4} ( 2 π ^{2} c ^{2} o ^{2} - 2 π co sin ( 2 π co ) - cos ( 2 π co ) + 1 )}{8 c ^{2} o ^{2}}

\frac{d}{d x} (e^{x} cot (y) + 2 y csc (y))

\int \frac{1}{2 ( x - 2 )} d x

\int_{0}^{1} x^{2} e^{x^{3} - 1} d x

\frac{\partial}{\partial x} (x^{2} + y^{3} + w^{4})

a re a y = ∣ 3 x ∣, y = x^{2} - 4