What is the integral of sin(6x)cos(9x) ?

The integral of sin(6x)cos(9x) is 1/2 (-1/15 cos(15x)+1/3 cos(3x))+C

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integral of sin(6x)cos(9x)

\int sin (6 x) cos (9 x) d x

\frac{1}{2} (- \frac{1}{15} cos (15 x) + \frac{1}{3} cos (3 x)) + C

Solution steps

\int sin (6 x) cos (9 x) d x

Use the following identity:

cos (t) sin (s) = \frac{sin ( s + t ) + sin ( s - t )}{2}

= \int \frac{sin ( 6 x + 9 x ) + sin ( 6 x - 9 x )}{2} 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 sin (6 x + 9 x) + sin (6 x - 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 sin (6 x + 9 x) d x + \int sin (6 x - 9 x) d x)

\int sin (6 x + 9 x) d x = - \frac{1}{15} cos (15 x)

\int sin (6 x - 9 x) d x = \frac{1}{3} cos (3 x)

= \frac{1}{2} (- \frac{1}{15} cos (15 x) + \frac{1}{3} cos (3 x))

Add a constant to the solution

= \frac{1}{2} (- \frac{1}{15} cos (15 x) + \frac{1}{3} cos (3 x)) + C

\frac{\partial}{\partial x} (3 y^{6} cos (3 x))

\int_{0}^{9} 162 - x^{2} d x

\frac{\partial}{\partial x} (2 e^{x y^{2}})

\int \frac{( 3 x - 1 )}{x x ^{2} - 4} d x

y^{^{''}} - y = 8 x e^{x}

What is the integral of sin(6x)cos(9x) ?
The integral of sin(6x)cos(9x) is 1/2 (-1/15 cos(15x)+1/3 cos(3x))+C