What is the integral of sin(3x)cos(2x) ?

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

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integral of sin(3x)cos(2x)

\int sin (3 x) cos (2 x) d x

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

Solution steps

\int sin (3 x) cos (2 x) d x

Use the following identity:

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

= \int \frac{sin ( 3 x + 2 x ) + sin ( 3 x - 2 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 (3 x + 2 x) + sin (3 x - 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 (3 x + 2 x) d x + \int sin (3 x - 2 x) d x)

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

\int sin (3 x - 2 x) d x = - cos (x)

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

Add a constant to the solution

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

\int \frac{sec ( \frac{1}{x} ) tan ( \frac{1}{x} )}{x ^{2}} d x

\frac{d}{d x} (\frac{94 x}{99 x - 94})

\frac{d}{d x} (\frac{( e ^{x} )}{x + 1})

\int cos (x + y + z) d z

\frac{\partial}{\partial x} (sin (π (2 x - 2 y)))

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