What is the integral of sin(3x)sin(6x) ?

The integral of sin(3x)sin(6x) is 1/2 (1/3 sin(3x)-1/9 sin(9x))+C

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integral of sin(3x)sin(6x)

\int sin (3 x) sin (6 x) d x

\frac{1}{2} (\frac{1}{3} sin (3 x) - \frac{1}{9} sin (9 x)) + C

Solution steps

\int sin (3 x) sin (6 x) d x

Rewrite using trig identities

= \int \frac{1}{2} (- cos (9 x) + cos (- 3 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 (9 x) + cos (- 3 x) d x

Simplify

- cos (9 x) + cos (- 3 x) : cos (3 x) - cos (9 x)

= \frac{1}{2} \cdot \int cos (3 x) - cos (9 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 (3 x) d x - \int cos (9 x) d x)

\int cos (3 x) d x = \frac{1}{3} sin (3 x)

\int cos (9 x) d x = \frac{1}{9} sin (9 x)

= \frac{1}{2} (\frac{1}{3} sin (3 x) - \frac{1}{9} sin (9 x))

Add a constant to the solution

= \frac{1}{2} (\frac{1}{3} sin (3 x) - \frac{1}{9} sin (9 x)) + C

[x^{2} (x^{2} + 8 x)]^{6}

x^{2} \frac{d y}{d x} = y^{2}

\frac{x ^{4}}{16} + \frac{1}{2 x ^{2}}

\int \frac{1}{16 - 4 x ^{2}} d x

\int \frac{1}{2 y + 1} d y

What is the integral of sin(3x)sin(6x) ?
The integral of sin(3x)sin(6x) is 1/2 (1/3 sin(3x)-1/9 sin(9x))+C