What is the integral of sin(8x)cos(4x) ?

The integral of sin(8x)cos(4x) is 1/2 (-1/12 cos(12x)-1/4 cos(4x))+C

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integral of sin(8x)cos(4x)

\int sin (8 x) cos (4 x) d x

\frac{1}{2} (- \frac{1}{12} cos (12 x) - \frac{1}{4} cos (4 x)) + C

Solution steps

\int sin (8 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 ( 8 x + 4 x ) + sin ( 8 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 (8 x + 4 x) + sin (8 x - 4 x) d x

Simplify

= \frac{1}{2} \cdot \int sin (12 x) + sin (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 (12 x) d x + \int sin (4 x) d x)

\int sin (12 x) d x = - \frac{1}{12} cos (12 x)

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

= \frac{1}{2} (- \frac{1}{12} cos (12 x) - \frac{1}{4} cos (4 x))

Add a constant to the solution

= \frac{1}{2} (- \frac{1}{12} cos (12 x) - \frac{1}{4} cos (4 x)) + C

f (x) = \frac{4 x}{x - 3}, at x = 5

(\frac{10}{9}) t^{2}

5 y^{^{''}} + y^{^{'}} = - 6 x

\frac{\partial}{\partial x} (y - 3 x)

x \to 0 lim (\frac{x}{sin ( π x )})

What is the integral of sin(8x)cos(4x) ?
The integral of sin(8x)cos(4x) is 1/2 (-1/12 cos(12x)-1/4 cos(4x))+C