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

解答

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

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

求解步骤

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

使用三角恒等式改写

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

提出常数:

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

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

使用三角恒等式改写

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

使用积分加法定则:

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

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

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

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

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

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

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