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

解

\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}}

解答ステップ

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

定数を除く:

\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

u置換積分法を適用する

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

定数を除く:

\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

部分積分を適用する

= 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}

簡素化

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}

限度を計算する:

\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})

簡素化

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