What is the integral of sin(2x)sin(6x) ?

The integral of sin(2x)sin(6x) is 1/2 (1/4 sin(4x)-1/8 sin(8x))+C

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integral of sin(2x)sin(6x)

\int sin (2 x) sin (6 x) d x

\frac{1}{2} (\frac{1}{4} sin (4 x) - \frac{1}{8} sin (8 x)) + C

Solution steps

\int sin (2 x) sin (6 x) d x

Rewrite using trig identities

= \int \frac{1}{2} (- cos (8 x) + cos (- 4 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 (8 x) + cos (- 4 x) d x

Simplify

- cos (8 x) + cos (- 4 x) : cos (4 x) - cos (8 x)

= \frac{1}{2} \cdot \int cos (4 x) - cos (8 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 (4 x) d x - \int cos (8 x) d x)

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

\int cos (8 x) d x = \frac{1}{8} sin (8 x)

= \frac{1}{2} (\frac{1}{4} sin (4 x) - \frac{1}{8} sin (8 x))

Add a constant to the solution

= \frac{1}{2} (\frac{1}{4} sin (4 x) - \frac{1}{8} sin (8 x)) + C

What is the integral of sin(2x)sin(6x) ?
The integral of sin(2x)sin(6x) is 1/2 (1/4 sin(4x)-1/8 sin(8x))+C