integral of sin(ln(14x))

解答

\int sin (ln (14 x)) d x

\frac{1}{2} x (sin (ln (14 x)) - cos (ln (14 x))) + C

求解步骤

\int sin (ln (14 x)) d x

使用换元积分法

= \int sin (ln (u)) \frac{1}{14} d u

提出常数:

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

= \frac{1}{14} \cdot \int sin (ln (u)) d u

使用分布积分法

= \frac{1}{14} (u sin (ln (u)) - \int cos (ln (u)) d u)

\int cos (ln (u)) d u = \frac{1}{2} u sin (ln (u)) + \frac{1}{2} u cos (ln (u))

= \frac{1}{14} (u sin (ln (u)) - (\frac{1}{2} u sin (ln (u)) + \frac{1}{2} u cos (ln (u))))

u = 14 x

代回

= \frac{1}{14} (14 x sin (ln (14 x)) - (\frac{1}{2} \cdot 14 x sin (ln (14 x)) + \frac{1}{2} \cdot 14 x cos (ln (14 x))))

化简

\frac{1}{14} (14 x sin (ln (14 x)) - (\frac{1}{2} \cdot 14 x sin (ln (14 x)) + \frac{1}{2} \cdot 14 x cos (ln (14 x)))) : \frac{1}{2} x (sin (ln (14 x)) - cos (ln (14 x)))

= \frac{1}{2} x (sin (ln (14 x)) - cos (ln (14 x)))

解答补常数

= \frac{1}{2} x (sin (ln (14 x)) - cos (ln (14 x))) + C

d er i v a t i v e (\frac{x}{2 ( x ^{2} - 1 ) ^{2} - 1})^{2}

\int_{0}^{30} 4 x (30 - x) d x

\frac{\partial}{\partial x} (x^{3} e^{4 x y})

y^{^{'}} = \frac{y}{t} - \frac{1}{\frac{y}{t}}

t an g e n t f (x) = x + \frac{2}{x}, at x = 2