integral of vsin^2(3v)

解答

\int v sin^{2} (3 v) d v

\frac{1}{72} (18 v^{2} - 6 v sin (6 v) - cos (6 v)) + C

求解步骤

\int v sin^{2} (3 v) d v

使用三角恒等式改写

= \int \frac{1}{2} v (1 - cos (6 v)) d v

提出常数:

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

= \frac{1}{2} \cdot \int v (1 - cos (6 v)) d v

使用分布积分法

= \frac{1}{2} (v (v - \frac{1}{6} sin (6 v)) - \int v - \frac{1}{6} sin (6 v) d v)

\int v - \frac{1}{6} sin (6 v) d v = \frac{v ^{2}}{2} + \frac{1}{36} cos (6 v)

= \frac{1}{2} (v (v - \frac{1}{6} sin (6 v)) - (\frac{v ^{2}}{2} + \frac{1}{36} cos (6 v)))

化简

\frac{1}{2} (v (v - \frac{1}{6} sin (6 v)) - (\frac{v ^{2}}{2} + \frac{1}{36} cos (6 v))) : \frac{1}{72} (18 v^{2} - 6 v sin (6 v) - cos (6 v))

= \frac{1}{72} (18 v^{2} - 6 v sin (6 v) - cos (6 v))

解答补常数

= \frac{1}{72} (18 v^{2} - 6 v sin (6 v) - cos (6 v)) + C

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