integral from 0 to pia of x^2sin^2(x)

解

\int_{0}^{πa} x^{2} sin^{2} (x) d x

\frac{π ^{3} a ^{3} sin ^{2} ( πa )}{3} + \frac{π ^{3} a ^{3} cos ( 2 πa )}{6} + \frac{- π ^{2} a ^{2} sin ( 2 πa ) - πa cos ( 2 πa )}{4} + \frac{1}{8} sin (2 πa)

解答ステップ

\int_{0}^{πa} x^{2} sin^{2} (x) d x

部分積分を適用する

= [\frac{1}{3} x^{3} sin^{2} (x) - \int \frac{1}{3} x^{3} sin (2 x) d x]_{0}^{πa}

\int \frac{1}{3} x^{3} sin (2 x) d x = - \frac{1}{6} x^{3} cos (2 x) + \frac{1}{8} (2 x^{2} sin (2 x) + 2 x cos (2 x) - sin (2 x))

= [\frac{1}{3} x^{3} sin^{2} (x) - (- \frac{1}{6} x^{3} cos (2 x) + \frac{1}{8} (2 x^{2} sin (2 x) + 2 x cos (2 x) - sin (2 x)))]_{0}^{πa}

拡張

\frac{1}{3} x^{3} sin^{2} (x) - (- \frac{1}{6} x^{3} cos (2 x) + \frac{1}{8} (2 x^{2} sin (2 x) + 2 x cos (2 x) - sin (2 x))) : \frac{1}{3} x^{3} sin^{2} (x) + \frac{1}{6} x^{3} cos (2 x) - \frac{1}{4} x^{2} sin (2 x) - \frac{1}{4} x cos (2 x) + \frac{1}{8} sin (2 x)

= [\frac{1}{3} x^{3} sin^{2} (x) + \frac{1}{6} x^{3} cos (2 x) - \frac{1}{4} x^{2} sin (2 x) - \frac{1}{4} x cos (2 x) + \frac{1}{8} sin (2 x)]_{0}^{πa}

限度を計算する:

\frac{π ^{3}}{3} a^{3} sin^{2} (πa) + \frac{π ^{3}}{6} a^{3} cos (2 πa) - \frac{π ^{2}}{4} a^{2} sin (2 πa) - \frac{π}{4} a cos (2 πa) + \frac{1}{8} sin (2 πa)

= \frac{π ^{3}}{3} a^{3} sin^{2} (πa) + \frac{π ^{3}}{6} a^{3} cos (2 πa) - \frac{π ^{2}}{4} a^{2} sin (2 πa) - \frac{π}{4} a cos (2 πa) + \frac{1}{8} sin (2 πa)

簡素化

= \frac{π ^{3} a ^{3} sin ^{2} ( πa )}{3} + \frac{π ^{3} a ^{3} cos ( 2 πa )}{6} + \frac{- π ^{2} a ^{2} sin ( 2 πa ) - πa cos ( 2 πa )}{4} + \frac{1}{8} sin (2 πa)