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

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

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

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

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

Solution steps

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

Use the following identity:

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

= \int \frac{sin ( 3 x + 6 x ) + sin ( 3 x - 6 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 (3 x + 6 x) + sin (3 x - 6 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 (3 x + 6 x) d x + \int sin (3 x - 6 x) d x)

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

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

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

Add a constant to the solution

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

\int_{1}^{2} e^{x} d x

\int x e^{- a x} d x

\frac{1}{( 1 + s ^{2} ) ( s ^{2} + 3 )}

\int_{0}^{1} ∣ x ∣ d x

f (x) = x + x, at x = 1

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