What is the integral of sin(7x)sin(9x) ?

The integral of sin(7x)sin(9x) is 1/2 (1/2 sin(2x)-1/16 sin(16x))+C

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integral of sin(7x)sin(9x)

\int sin (7 x) sin (9 x) d x

\frac{1}{2} (\frac{1}{2} sin (2 x) - \frac{1}{16} sin (16 x)) + C

Solution steps

\int sin (7 x) sin (9 x) d x

Rewrite using trig identities

= \int \frac{1}{2} (- cos (16 x) + cos (- 2 x)) d x

Take the constant out:

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

= \frac{1}{2} \cdot \int - cos (16 x) + cos (- 2 x) d x

Simplify

- cos (16 x) + cos (- 2 x) : cos (2 x) - cos (16 x)

= \frac{1}{2} \cdot \int cos (2 x) - cos (16 x) d x

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 x) d x - \int cos (16 x) d x)

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

\int cos (16 x) d x = \frac{1}{16} sin (16 x)

= \frac{1}{2} (\frac{1}{2} sin (2 x) - \frac{1}{16} sin (16 x))

Add a constant to the solution

= \frac{1}{2} (\frac{1}{2} sin (2 x) - \frac{1}{16} sin (16 x)) + C

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What is the integral of sin(7x)sin(9x) ?
The integral of sin(7x)sin(9x) is 1/2 (1/2 sin(2x)-1/16 sin(16x))+C