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

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

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

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

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

Solution steps

\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}

= \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

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

Simplify

= \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

= \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)

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

Add a constant to the solution

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

\int sec^{2} (x) tan^{7} (x) d x

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

e^{x^{5}}

f (x) = x^{3} - 3 x^{2}

\frac{d ^{2}}{d x ^{2}} (x + 3 x)

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