integral of sin(ln(u))

解答

\int sin (ln (u)) d u

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

求解步骤

\int sin (ln (u)) d u

使用分布积分法

= 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))

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

化简

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

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

解答补常数

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

y^{^{''}} - y = 5 sin (2 t) (0) 1

t a y l or \frac{1}{2 + 3 x}

y^{^{'}} = y + x, y (0) = - 1

\int (- 0.5 x + 450) d x

\frac{d y}{d x} = 4 x - 5