What is the integral of 3sin(5x)cos(4x) ?

The integral of 3sin(5x)cos(4x) is 3/2 (-1/9 cos(9x)-cos(x))+C

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integral of 3sin(5x)cos(4x)

\int 3 sin (5 x) cos (4 x) d x

\frac{3}{2} (- \frac{1}{9} cos (9 x) - cos (x)) + C

Solution steps

\int 3 sin (5 x) cos (4 x) d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 3 \cdot \int sin (5 x) cos (4 x) d x

Use the following identity:

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

= 3 \cdot \int \frac{sin ( 5 x + 4 x ) + sin ( 5 x - 4 x )}{2} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 3 \cdot \frac{1}{2} \cdot \int sin (5 x + 4 x) + sin (5 x - 4 x) d x

Simplify

= 3 \cdot \frac{1}{2} \cdot \int sin (9 x) + sin (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

= 3 \cdot \frac{1}{2} (\int sin (9 x) d x + \int sin (x) d x)

\int sin (9 x) d x = - \frac{1}{9} cos (9 x)

\int sin (x) d x = - cos (x)

= 3 \cdot \frac{1}{2} (- \frac{1}{9} cos (9 x) - cos (x))

Simplify

3 \cdot \frac{1}{2} (- \frac{1}{9} cos (9 x) - cos (x)) : \frac{3}{2} (- \frac{1}{9} cos (9 x) - cos (x))

= \frac{3}{2} (- \frac{1}{9} cos (9 x) - cos (x))

Add a constant to the solution

= \frac{3}{2} (- \frac{1}{9} cos (9 x) - cos (x)) + C

What is the integral of 3sin(5x)cos(4x) ?
The integral of 3sin(5x)cos(4x) is 3/2 (-1/9 cos(9x)-cos(x))+C