integral from 0 to t of (t-x)^2cos(2x)

解

\int_{0}^{t} (t - x)^{2} cos (2 x) d x

\frac{t ^{2} sin ( 2 t )}{2} - \frac{t ( 2 t sin ( 2 t ) + cos ( 2 t ) - 1 )}{2} + \frac{t ^{2} sin ( 2 t ) + t cos ( 2 t )}{2} - \frac{1}{4} sin (2 t)

解答ステップ

\int_{0}^{t} (t - x)^{2} cos (2 x) d x

拡張

(t - x)^{2} cos (2 x) : t^{2} cos (2 x) - 2 t x cos (2 x) + x^{2} cos (2 x)

= \int_{0}^{t} t^{2} cos (2 x) - 2 t x cos (2 x) + x^{2} cos (2 x) d x

総和規則を適用する:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{0}^{t} t^{2} cos (2 x) d x - \int_{0}^{t} 2 t x cos (2 x) d x + \int_{0}^{t} x^{2} cos (2 x) d x

\int_{0}^{t} t^{2} cos (2 x) d x = \frac{t ^{2} sin ( 2 t )}{2}

\int_{0}^{t} 2 t x cos (2 x) d x = \frac{t ( 2 t sin ( 2 t ) + cos ( 2 t ) - 1 )}{2}

\int_{0}^{t} x^{2} cos (2 x) d x = \frac{t ^{2} sin ( 2 t ) + t cos ( 2 t )}{2} - \frac{1}{4} sin (2 t)

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