integral from 0 to 2 of sin^2(xco)s^22x

解答

\int_{0}^{2} sin^{2} (x co) s^{2} 2 x

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

求解步骤

\int_{0}^{2} sin^{2} (x co) s^{2} \cdot 2 x d x

提出常数:

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

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

使用换元积分法

= 2 s^{2} \cdot \int_{0}^{co \cdot 2} \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

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

使用分布积分法

= 2 s^{2} \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}^{2 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))

= 2 s^{2} \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}^{2 co}

化简

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

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

带入上下限计算后得:

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

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

化简

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

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