integral from 0 to pi of cos(2t)+sin(2t)

解答

\int_{0}^{π} cos (2 t) + sin (2 t) d t

0

求解步骤

\int_{0}^{π} cos (2 t) + sin (2 t) d t

使用三角恒等式改写

= \int_{0}^{π} - 1 + 2 cos^{2} (t) + 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

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

\int_{0}^{π} 1 d t = π

\int_{0}^{π} 2 cos^{2} (t) d t = π

\int_{0}^{π} sin (2 t) d t = 0

= - π + π + 0

化简

= 0

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