What is the integral of sin(ln(z)) ?

The integral of sin(ln(z)) is 1/2 zsin(ln(z))-1/2 zcos(ln(z))+C

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integral of sin(ln(z))

\int sin (ln (z)) d z

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

Solution steps

\int sin (ln (z)) d z

Apply Integration By Parts

: z sin (ln (z)) - \int cos (ln (z)) d z

= z sin (ln (z)) - \int cos (ln (z)) d z

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

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

Simplify

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

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

Add a constant to the solution

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

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What is the integral of sin(ln(z)) ?
The integral of sin(ln(z)) is 1/2 zsin(ln(z))-1/2 zcos(ln(z))+C