What is the integral of zsin(z)cos(z) ?

The integral of zsin(z)cos(z) is 1/2 (-1/2 zcos(2z)+1/4 sin(2z))+C

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integral of zsin(z)cos(z)

\int z sin (z) cos (z) d z

\frac{1}{2} (- \frac{1}{2} z cos (2 z) + \frac{1}{4} sin (2 z)) + C

Solution steps

\int z sin (z) cos (z) d z

Rewrite using trig identities

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

Take the constant out:

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

= \frac{1}{2} \cdot \int z sin (2 z) d z

Apply Integration By Parts

: - \frac{1}{2} z cos (2 z) - \int - \frac{1}{2} cos (2 z) d z

= \frac{1}{2} (- \frac{1}{2} z cos (2 z) - \int - \frac{1}{2} cos (2 z) d z)

\int - \frac{1}{2} cos (2 z) d z = - \frac{1}{4} sin (2 z)

= \frac{1}{2} (- \frac{1}{2} z cos (2 z) - (- \frac{1}{4} sin (2 z)))

Simplify

= \frac{1}{2} (- \frac{1}{2} z cos (2 z) + \frac{1}{4} sin (2 z))

Add a constant to the solution

= \frac{1}{2} (- \frac{1}{2} z cos (2 z) + \frac{1}{4} sin (2 z)) + C

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What is the integral of zsin(z)cos(z) ?
The integral of zsin(z)cos(z) is 1/2 (-1/2 zcos(2z)+1/4 sin(2z))+C