What is the integral of sin(8x)cos(3x) ?

The integral of sin(8x)cos(3x) is 1/2 (-1/11 cos(11x)-1/5 cos(5x))+C

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integral of sin(8x)cos(3x)

\int sin (8 x) cos (3 x) d x

\frac{1}{2} (- \frac{1}{11} cos (11 x) - \frac{1}{5} cos (5 x)) + C

Solution steps

\int sin (8 x) cos (3 x) d x

Use the following identity:

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

= \int \frac{sin ( 8 x + 3 x ) + sin ( 8 x - 3 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 + 3 x) + sin (8 x - 3 x) d x

Simplify

= \frac{1}{2} \cdot \int sin (11 x) + sin (5 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 (11 x) d x + \int sin (5 x) d x)

\int sin (11 x) d x = - \frac{1}{11} cos (11 x)

\int sin (5 x) d x = - \frac{1}{5} cos (5 x)

= \frac{1}{2} (- \frac{1}{11} cos (11 x) - \frac{1}{5} cos (5 x))

Add a constant to the solution

= \frac{1}{2} (- \frac{1}{11} cos (11 x) - \frac{1}{5} cos (5 x)) + C

\int_{- 1}^{4} x + 1 d x

\frac{1}{x ln ( x )}

\int_{0}^{π} 17 sin^{2} (4 x) d x

y^{^{'''}} + 4 y^{^{'}} = 3 x - 1

What is the integral of sin(8x)cos(3x) ?
The integral of sin(8x)cos(3x) is 1/2 (-1/11 cos(11x)-1/5 cos(5x))+C