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受欢迎的微积分 >

integral from 0 to pi of sin^3(xco)s^2x

解答

\int_{0}^{π} sin^{3} (x co) s^{2} x

解答

\frac{s ^{2} ( 3 π co cos ^{3} ( π co ) - 9 π co cos ( π co ) + sin ^{3} ( π co ) + 6 sin ( π co ) )}{9 c ^{2} o ^{2}}

求解步骤

\int_{0}^{π} sin^{3} (x co) s^{2} x d x

提出常数:

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

= s^{2} \cdot \int_{0}^{π} sin^{3} (co x) x d x

使用换元积分法

= s^{2} \cdot \int_{0}^{co π} \frac{u sin ^{3} ( 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

= s^{2} \frac{1}{c ^{2} o ^{2}} \cdot \int_{0}^{co π} u sin^{3} (u) d u

使用分布积分法

= s^{2} \frac{1}{c ^{2} o ^{2}} [u (- cos (u) + \frac{1}{3} cos^{3} (u)) - \int - cos (u) + \frac{1}{3} cos^{3} (u) d u]_{0}^{π co}

\int - cos (u) + \frac{1}{3} cos^{3} (u) d u = - \frac{2}{3} sin (u) - \frac{1}{9} sin^{3} (u)

= s^{2} \frac{1}{c ^{2} o ^{2}} [u (- cos (u) + \frac{1}{3} cos^{3} (u)) - (- \frac{2}{3} sin (u) - \frac{1}{9} sin^{3} (u))]_{0}^{π co}

乘开

u (- cos (u) + \frac{1}{3} cos^{3} (u)) - (- \frac{2}{3} sin (u) - \frac{1}{9} sin^{3} (u)) : - u cos (u) + \frac{1}{3} u cos^{3} (u) + \frac{2}{3} sin (u) + \frac{1}{9} sin^{3} (u)

= s^{2} \frac{1}{c ^{2} o ^{2}} [- u cos (u) + \frac{1}{3} u cos^{3} (u) + \frac{2}{3} sin (u) + \frac{1}{9} sin^{3} (u)]_{0}^{π co}

带入上下限计算后得:

\frac{π}{3} co cos^{3} (π co) - π co cos (π co) + \frac{1}{9} sin^{3} (π co) + \frac{2}{3} sin (π co)

= s^{2} \frac{1}{c ^{2} o ^{2}} (\frac{π}{3} co cos^{3} (π co) - π co cos (π co) + \frac{1}{9} sin^{3} (π co) + \frac{2}{3} sin (π co))

化简

= \frac{s ^{2} ( 3 π co cos ^{3} ( π co ) - 9 π co cos ( π co ) + sin ^{3} ( π co ) + 6 sin ( π co ) )}{9 c ^{2} o ^{2}}

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