integral from 0 to pi of 9sin^2(xco)s^4x

解答

\int_{0}^{π} 9 sin^{2} (x co) s^{4} x d x

\frac{9 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} 9 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

= 9 s^{4} \cdot \int_{0}^{π} sin^{2} (co x) x d x

使用换元积分法

= 9 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

= 9 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

= 9 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))

= 9 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

9 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{9 s ^{4}}{c ^{2} o ^{2}} [\frac{2 u ^{2} - 2 u sin ( 2 u ) - cos ( 2 u )}{8}]_{0}^{π co}

= \frac{9 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{9 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{9 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^{s i n (5 x)})

\int cos^{4} (2 y) d y

\frac{\partial}{\partial x} (e^{- 2 x} cos (2 π t))

y^{^{'}} = \frac{( 4 - 2 x )}{( 3 + 2 y )}

t an g e n t \frac{3 x}{x ^{2} - 6}