integral from 0 to t of sin(x)sin(t-x)

解

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

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

解答ステップ

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

三角関数の公式を使用して書き換える

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

定数を除く:

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

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

三角関数の公式を使用して書き換える

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

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

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

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

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

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