integral from pi to x of t^3sin^2(t)

解

\int_{π}^{x} t^{3} sin^{2} (t) d t

\frac{x ^{4} sin ^{2} ( x ) - x ^{3} sin ( 2 x )}{4} + \frac{x ^{4} cos ( 2 x ) - 3 x ^{2} cos ( 2 x ) + 3 x sin ( 2 x )}{8} + \frac{3}{16} cos (2 x) + \frac{- 2 π ^{4} - 3 ( - 2 π ^{2} + 1 )}{16}

解答ステップ

\int_{π}^{x} t^{3} sin^{2} (t) d t

部分積分を適用する

= [\frac{1}{4} t^{4} sin^{2} (t) - \int \frac{1}{4} t^{4} sin (2 t) d t]_{π}^{x}

\int \frac{1}{4} t^{4} sin (2 t) d t = \frac{1}{16} (- 2 t^{4} cos (2 t) + 4 t^{3} sin (2 t) - 3 (- 2 t^{2} cos (2 t) + 2 t sin (2 t) + cos (2 t)))

= [\frac{1}{4} t^{4} sin^{2} (t) - \frac{1}{16} (- 2 t^{4} cos (2 t) + 4 t^{3} sin (2 t) - 3 (- 2 t^{2} cos (2 t) + 2 t sin (2 t) + cos (2 t)))]_{π}^{x}

限度を計算する:

\frac{1}{4} x^{4} sin^{2} (x) - \frac{1}{16} (- 2 x^{4} cos (2 x) + 4 x^{3} sin (2 x) - 3 (- 2 x^{2} cos (2 x) + 2 x sin (2 x) + cos (2 x))) + \frac{- 2 π ^{4} - 3 ( - 2 π ^{2} + 1 )}{16}

= \frac{1}{4} x^{4} sin^{2} (x) - \frac{1}{16} (- 2 x^{4} cos (2 x) + 4 x^{3} sin (2 x) - 3 (- 2 x^{2} cos (2 x) + 2 x sin (2 x) + cos (2 x))) + \frac{- 2 π ^{4} - 3 ( - 2 π ^{2} + 1 )}{16}

簡素化

= \frac{x ^{4} sin ^{2} ( x ) - x ^{3} sin ( 2 x )}{4} + \frac{x ^{4} cos ( 2 x ) - 3 x ^{2} cos ( 2 x ) + 3 x sin ( 2 x )}{8} + \frac{3}{16} cos (2 x) + \frac{- 2 π ^{4} - 3 ( - 2 π ^{2} + 1 )}{16}