What is the integral of sin(5z)sin(7z) ?

The integral of sin(5z)sin(7z) is 1/2 (1/2 sin(2z)-1/12 sin(12z))+C

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integral of sin(5z)sin(7z)

\int sin (5 z) sin (7 z) d z

\frac{1}{2} (\frac{1}{2} sin (2 z) - \frac{1}{12} sin (12 z)) + C

Solution steps

\int sin (5 z) sin (7 z) d z

Rewrite using trig identities

= \int \frac{1}{2} (- cos (12 z) + cos (- 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 - cos (12 z) + cos (- 2 z) d z

Simplify

- cos (12 z) + cos (- 2 z) : cos (2 z) - cos (12 z)

= \frac{1}{2} \cdot \int cos (2 z) - cos (12 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 (2 z) d z - \int cos (12 z) d z)

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

\int cos (12 z) d z = \frac{1}{12} sin (12 z)

= \frac{1}{2} (\frac{1}{2} sin (2 z) - \frac{1}{12} sin (12 z))

Add a constant to the solution

= \frac{1}{2} (\frac{1}{2} sin (2 z) - \frac{1}{12} sin (12 z)) + C

What is the integral of sin(5z)sin(7z) ?
The integral of sin(5z)sin(7z) is 1/2 (1/2 sin(2z)-1/12 sin(12z))+C