What is the integral of sin(2x)cos(6x) ?

The integral of sin(2x)cos(6x) is 1/2 (-1/8 cos(8x)+1/4 cos(4x))+C

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integral of sin(2x)cos(6x)

\int sin (2 x) cos (6 x) d x

\frac{1}{2} (- \frac{1}{8} cos (8 x) + \frac{1}{4} cos (4 x)) + C

Solution steps

\int sin (2 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 ( 2 x + 6 x ) + sin ( 2 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 (2 x + 6 x) + sin (2 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 (2 x + 6 x) d x + \int sin (2 x - 6 x) d x)

\int sin (2 x + 6 x) d x = - \frac{1}{8} cos (8 x)

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

= \frac{1}{2} (- \frac{1}{8} cos (8 x) + \frac{1}{4} cos (4 x))

Add a constant to the solution

= \frac{1}{2} (- \frac{1}{8} cos (8 x) + \frac{1}{4} cos (4 x)) + C

\int 2 x sin (4 x) d x

x \to \infty lim ((x)^{2})

\frac{d}{d x} (\frac{x - 22}{x - 22})

\frac{\partial}{\partial x} (\frac{a}{x})

\frac{d}{d y} (\frac{x ^{2} y}{x ^{2} + y ^{2}})

What is the integral of sin(2x)cos(6x) ?
The integral of sin(2x)cos(6x) is 1/2 (-1/8 cos(8x)+1/4 cos(4x))+C