What is the integral of 9sin(4x)cos(2x) ?

The integral of 9sin(4x)cos(2x) is 9/2 (-1/6 cos(6x)-1/2 cos(2x))+C

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integral of 9sin(4x)cos(2x)

\int 9 sin (4 x) cos (2 x) d x

\frac{9}{2} (- \frac{1}{6} cos (6 x) - \frac{1}{2} cos (2 x)) + C

Solution steps

\int 9 sin (4 x) cos (2 x) d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 9 \cdot \int sin (4 x) cos (2 x) d x

Use the following identity:

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

= 9 \cdot \int \frac{sin ( 4 x + 2 x ) + sin ( 4 x - 2 x )}{2} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 9 \cdot \frac{1}{2} \cdot \int sin (4 x + 2 x) + sin (4 x - 2 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

= 9 \cdot \frac{1}{2} (\int sin (4 x + 2 x) d x + \int sin (4 x - 2 x) d x)

\int sin (4 x + 2 x) d x = - \frac{1}{6} cos (6 x)

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

= 9 \cdot \frac{1}{2} (- \frac{1}{6} cos (6 x) - \frac{1}{2} cos (2 x))

Simplify

9 \cdot \frac{1}{2} (- \frac{1}{6} cos (6 x) - \frac{1}{2} cos (2 x)) : \frac{9}{2} (- \frac{1}{6} cos (6 x) - \frac{1}{2} cos (2 x))

= \frac{9}{2} (- \frac{1}{6} cos (6 x) - \frac{1}{2} cos (2 x))

Add a constant to the solution

= \frac{9}{2} (- \frac{1}{6} cos (6 x) - \frac{1}{2} cos (2 x)) + C

x^{^{'''}} + x = 0

y^{^{'}} = - sin (π x)

\frac{d}{d x} (ln (- 3 x + 2))

\int_{0}^{- 0.4} 150 x d x

\int x sin (\frac{x}{8}) d x

What is the integral of 9sin(4x)cos(2x) ?
The integral of 9sin(4x)cos(2x) is 9/2 (-1/6 cos(6x)-1/2 cos(2x))+C