What is the integral of zcos(3z) ?

The integral of zcos(3z) is 1/9 (3zsin(3z)+cos(3z))+C

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integral of zcos(3z)

\int z cos (3 z) d z

\frac{1}{9} (3 z sin (3 z) + cos (3 z)) + C

Solution steps

\int z cos (3 z) d z

Apply u-substitution

: \int \frac{u cos ( u )}{9} d u

= \int \frac{u cos ( u )}{9} d u

Take the constant out:

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

= \frac{1}{9} \cdot \int u cos (u) d u

Apply Integration By Parts

: u sin (u) - \int sin (u) d u

= \frac{1}{9} (u sin (u) - \int sin (u) d u)

\int sin (u) d u = - cos (u)

= \frac{1}{9} (u sin (u) - (- cos (u)))

Substitute back

u = 3 z

= \frac{1}{9} (3 z sin (3 z) - (- cos (3 z)))

Simplify

= \frac{1}{9} (3 z sin (3 z) + cos (3 z))

Add a constant to the solution

= \frac{1}{9} (3 z sin (3 z) + cos (3 z)) + C

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What is the integral of zcos(3z) ?
The integral of zcos(3z) is 1/9 (3zsin(3z)+cos(3z))+C