What is the integral of sin(13z)sin(9z) ?

The integral of sin(13z)sin(9z) is 1/2 (-1/22 sin(22z)+1/4 sin(4z))+C

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integral of sin(13z)sin(9z)

\int sin (13 z) sin (9 z) d z

\frac{1}{2} (- \frac{1}{22} sin (22 z) + \frac{1}{4} sin (4 z)) + C

Solution steps

\int sin (13 z) sin (9 z) d z

Rewrite using trig identities

= \int \frac{1}{2} (- cos (22 z) + cos (4 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 - cos (22 z) + cos (4 z) d z

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} (- \int cos (22 z) d z + \int cos (4 z) d z)

\int cos (22 z) d z = \frac{1}{22} sin (22 z)

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

= \frac{1}{2} (- \frac{1}{22} sin (22 z) + \frac{1}{4} sin (4 z))

Add a constant to the solution

= \frac{1}{2} (- \frac{1}{22} sin (22 z) + \frac{1}{4} sin (4 z)) + C

What is the integral of sin(13z)sin(9z) ?
The integral of sin(13z)sin(9z) is 1/2 (-1/22 sin(22z)+1/4 sin(4z))+C