integral from 0 to pi/4 of t^2sin(4t)

解

\int_{0}^{\frac{π}{4}} t^{2} sin (4 t) d t

\frac{π ^{2} - 2}{64} - \frac{1}{32}

十進法表記

0.09171 \dots

解答ステップ

\int_{0}^{\frac{π}{4}} t^{2} sin (4 t) d t

部分積分を適用する

= [\frac{1}{4} (- t^{2} cos (4 t) - 4 \cdot \int - \frac{1}{2} t cos (4 t) d t)]_{0}^{\frac{π}{4}}

\int - \frac{1}{2} t cos (4 t) d t = - \frac{1}{32} (4 t sin (4 t) + cos (4 t))

= [\frac{1}{4} (- t^{2} cos (4 t) - 4 (- \frac{1}{32} (4 t sin (4 t) + cos (4 t))))]_{0}^{\frac{π}{4}}

簡素化

[\frac{1}{4} (- t^{2} cos (4 t) - 4 (- \frac{1}{32} (4 t sin (4 t) + cos (4 t))))]_{0}^{\frac{π}{4}} : [\frac{1}{4} (- t^{2} cos (4 t) + \frac{1}{8} (4 t sin (4 t) + cos (4 t)))]_{0}^{\frac{π}{4}}

= [\frac{1}{4} (- t^{2} cos (4 t) + \frac{1}{8} (4 t sin (4 t) + cos (4 t)))]_{0}^{\frac{π}{4}}

限度を計算する:

\frac{π ^{2} - 2}{64} - \frac{1}{32}

= \frac{π ^{2} - 2}{64} - \frac{1}{32}