integral from 0 to pi/2 of x^2ycos(xy^2)

解

\int_{0}^{\frac{π}{2}} x^{2} y cos (x y^{2}) d x

\frac{π ^{2} y ^{4} sin ( \frac{π y ^{2}}{2} ) - 8 sin ( \frac{π y ^{2}}{2} ) + 4 π y ^{2} cos ( \frac{π y ^{2}}{2} )}{4 y ^{5}}

解答ステップ

\int_{0}^{\frac{π}{2}} x^{2} y cos (x y^{2}) d x

定数を除く:

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

= y \cdot \int_{0}^{\frac{π}{2}} cos (y^{2} x) x^{2} d x

簡素化

y^{2} : yy

= y \cdot \int_{0}^{\frac{π}{2}} cos (yy x) x^{2} d x

u置換積分法を適用する

= y \cdot \int_{0}^{\frac{π y ^{2}}{2}} \frac{u ^{2} cos ( u )}{y ^{6}} d u

定数を除く:

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

= y \frac{1}{y ^{6}} \cdot \int_{0}^{\frac{π y ^{2}}{2}} u^{2} cos (u) d u

部分積分を適用する

= y \frac{1}{y ^{6}} [u^{2} sin (u) - \int 2 u sin (u) d u]_{0}^{\frac{π y ^{2}}{2}}

\int 2 u sin (u) d u = 2 (- u cos (u) + sin (u))

= y \frac{1}{y ^{6}} [u^{2} sin (u) - 2 (- u cos (u) + sin (u))]_{0}^{\frac{π y ^{2}}{2}}

簡素化

y \frac{1}{y ^{6}} [u^{2} sin (u) - 2 (- u cos (u) + sin (u))]_{0}^{\frac{π y ^{2}}{2}} : \frac{1}{y ^{5}} [u^{2} sin (u) - 2 (- u cos (u) + sin (u))]_{0}^{\frac{π y ^{2}}{2}}

= \frac{1}{y ^{5}} [u^{2} sin (u) - 2 (- u cos (u) + sin (u))]_{0}^{\frac{π y ^{2}}{2}}

限度を計算する:

\frac{π ^{2} y ^{4}}{4} sin (\frac{π y ^{2}}{2}) - 2 sin (\frac{π y ^{2}}{2}) + π y^{2} cos (\frac{π y ^{2}}{2})

= \frac{1}{y ^{5}} (\frac{π ^{2} y ^{4}}{4} sin (\frac{π y ^{2}}{2}) - 2 sin (\frac{π y ^{2}}{2}) + π y^{2} cos (\frac{π y ^{2}}{2}))

簡素化

= \frac{π ^{2} y ^{4} sin ( \frac{π y ^{2}}{2} ) - 8 sin ( \frac{π y ^{2}}{2} ) + 4 π y ^{2} cos ( \frac{π y ^{2}}{2} )}{4 y ^{5}}