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

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

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

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

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

Solution steps

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

Use the following identity:

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

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

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

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

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

Add a constant to the solution

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

\int_{1}^{3} 4 x e^{x} d x

x \to 1 lim (π)

sin (π x)

t \to 0 lim (\frac{7}{1 + t})

\frac{\partial}{\partial y} (arctan (x + 8 y))

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