What is the integral of sin(8x)cos(6x) ?

The integral of sin(8x)cos(6x) is 1/2 (-1/14 cos(14x)-1/2 cos(2x))+C

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integral of sin(8x)cos(6x)

\int sin (8 x) cos (6 x) d x

\frac{1}{2} (- \frac{1}{14} cos (14 x) - \frac{1}{2} cos (2 x)) + C

Solution steps

\int sin (8 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 ( 8 x + 6 x ) + sin ( 8 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 (8 x + 6 x) + sin (8 x - 6 x) d x

Simplify

= \frac{1}{2} \cdot \int sin (14 x) + sin (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

= \frac{1}{2} (\int sin (14 x) d x + \int sin (2 x) d x)

\int sin (14 x) d x = - \frac{1}{14} cos (14 x)

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

= \frac{1}{2} (- \frac{1}{14} cos (14 x) - \frac{1}{2} cos (2 x))

Add a constant to the solution

= \frac{1}{2} (- \frac{1}{14} cos (14 x) - \frac{1}{2} cos (2 x)) + C

(2 ln (x))^{^{'}}

\int (x + 5)^{7} d x

\int_{1}^{2} x x - 1 d x

\frac{d}{d x} (a x^{4})

\frac{\partial}{\partial x} (\frac{( 5 x + 4 )}{5 y + 1})

What is the integral of sin(8x)cos(6x) ?
The integral of sin(8x)cos(6x) is 1/2 (-1/14 cos(14x)-1/2 cos(2x))+C