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

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

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

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

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

Solution steps

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

Use the following identity:

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

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

Simplify

= \frac{1}{2} \cdot \int sin (9 x) + sin (3 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 (9 x) d x + \int sin (3 x) d x)

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

\int sin (3 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

\frac{d}{d x} ((x)^{x + 1})

\frac{d x}{d y} = sin (x + y)

(2 x - 1)^{2} \cdot ln (2 x - 1)

(0.4) (6.8)

y^{^{'}} = - a y

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