integral of sin(ln(6x))

解答

\int sin (ln (6 x)) d x

\frac{1}{2} x (sin (ln (6 x)) - cos (ln (6 x))) + C

求解步骤

\int sin (ln (6 x)) d x

使用换元积分法

= \int sin (ln (u)) \frac{1}{6} d u

提出常数:

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

= \frac{1}{6} \cdot \int sin (ln (u)) d u

使用分布积分法

= \frac{1}{6} (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}{6} (u sin (ln (u)) - (\frac{1}{2} u sin (ln (u)) + \frac{1}{2} u cos (ln (u))))

u = 6 x

代回

= \frac{1}{6} (6 x sin (ln (6 x)) - (\frac{1}{2} \cdot 6 x sin (ln (6 x)) + \frac{1}{2} \cdot 6 x cos (ln (6 x))))

化简

\frac{1}{6} (6 x sin (ln (6 x)) - (\frac{1}{2} \cdot 6 x sin (ln (6 x)) + \frac{1}{2} \cdot 6 x cos (ln (6 x)))) : \frac{1}{2} x (sin (ln (6 x)) - cos (ln (6 x)))

= \frac{1}{2} x (sin (ln (6 x)) - cos (ln (6 x)))

解答补常数

= \frac{1}{2} x (sin (ln (6 x)) - cos (ln (6 x))) + C

\frac{d}{d x} (2 \cdot 3 x)

d er i v a t i v e f (x) = - 4 x + 8

n = 5 \sum \infty \frac{7 ^{n}}{8 ^{n}}

\frac{d y}{d x} = 7^{x} ln (7) + \frac{2}{x ^{2} + 1}

\frac{d ^{3}}{d x ^{3}} (\frac{10}{x ^{2}})