integral of xsin^2(8x)

解答

\int x sin^{2} (8 x) d x

\frac{1}{512} (128 x^{2} - 16 x sin (16 x) - cos (16 x)) + C

求解步骤

\int x sin^{2} (8 x) d x

使用三角恒等式改写

= \int \frac{1}{2} x (1 - cos (16 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 (16 x)) d x

使用分布积分法

= \frac{1}{2} (x (x - \frac{1}{16} sin (16 x)) - \int x - \frac{1}{16} sin (16 x) d x)

\int x - \frac{1}{16} sin (16 x) d x = \frac{x ^{2}}{2} + \frac{1}{256} cos (16 x)

= \frac{1}{2} (x (x - \frac{1}{16} sin (16 x)) - (\frac{x ^{2}}{2} + \frac{1}{256} cos (16 x)))

化简

\frac{1}{2} (x (x - \frac{1}{16} sin (16 x)) - (\frac{x ^{2}}{2} + \frac{1}{256} cos (16 x))) : \frac{1}{512} (128 x^{2} - 16 x sin (16 x) - cos (16 x))

= \frac{1}{512} (128 x^{2} - 16 x sin (16 x) - cos (16 x))

解答补常数

= \frac{1}{512} (128 x^{2} - 16 x sin (16 x) - cos (16 x)) + C

\int 3 e^{x} sin (x) d x

x \to 0 lim (\frac{3 ^{s i n (x)} - 1}{4 x})

d er i v a t i v e e^{x^{4}}

\int t^{2} cos^{2} (t) d t

\int \frac{x + 10}{2 - x - 3 x ^{2}} d x