integral of xsin^2(6x)

解答

\int x sin^{2} (6 x) d x

\frac{1}{288} (72 x^{2} - 12 x sin (12 x) - cos (12 x)) + C

求解步骤

\int x sin^{2} (6 x) d x

使用三角恒等式改写

= \int \frac{1}{2} x (1 - cos (12 x)) d x

提出常数:

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

= \frac{1}{2} \cdot \int x (1 - cos (12 x)) d x

使用分布积分法

= \frac{1}{2} (x (x - \frac{1}{12} sin (12 x)) - \int x - \frac{1}{12} sin (12 x) d x)

\int x - \frac{1}{12} sin (12 x) d x = \frac{x ^{2}}{2} + \frac{1}{144} cos (12 x)

= \frac{1}{2} (x (x - \frac{1}{12} sin (12 x)) - (\frac{x ^{2}}{2} + \frac{1}{144} cos (12 x)))

化简

\frac{1}{2} (x (x - \frac{1}{12} sin (12 x)) - (\frac{x ^{2}}{2} + \frac{1}{144} cos (12 x))) : \frac{1}{288} (72 x^{2} - 12 x sin (12 x) - cos (12 x))

= \frac{1}{288} (72 x^{2} - 12 x sin (12 x) - cos (12 x))

解答补常数

= \frac{1}{288} (72 x^{2} - 12 x sin (12 x) - cos (12 x)) + C

\frac{d}{d x} (4 x^{3} - 46 x^{2} + 120 x)

\int \frac{1}{x} d x

\int sin (5 x) cos (x) d x

\int \frac{x ^{3} - x ^{2} + 2 x - 6}{x} d x

a re a 4 x + 7, 2 x + 29, 6, 16