What is the integral of ((sin(3x)-cos(2x)))/4 ?

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

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integral of ((sin(3x)-cos(2x)))/4

\int \frac{( sin ( 3 x ) - cos ( 2 x ) )}{4} d x

\frac{1}{4} (- \frac{1}{3} cos (3 x) - \frac{1}{2} sin (2 x)) + C

Solution steps

\int \frac{sin ( 3 x ) - cos ( 2 x )}{4} d x

Take the constant out:

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

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

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

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

= \frac{1}{4} (- \frac{1}{3} cos (3 x) - \frac{1}{2} sin (2 x))

Add a constant to the solution

= \frac{1}{4} (- \frac{1}{3} cos (3 x) - \frac{1}{2} sin (2 x)) + C

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