What is the integral of 4sin(2x)-3cos(6x) ?

The integral of 4sin(2x)-3cos(6x) is -2cos(2x)-1/2 sin(6x)+C

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integral of 4sin(2x)-3cos(6x)

\int 4 sin (2 x) - 3 cos (6 x) d x

- 2 cos (2 x) - \frac{1}{2} sin (6 x) + C

Solution steps

\int 4 sin (2 x) - 3 cos (6 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

= \int 4 sin (2 x) d x - \int 3 cos (6 x) d x

\int 4 sin (2 x) d x = - 2 cos (2 x)

\int 3 cos (6 x) d x = \frac{1}{2} sin (6 x)

= - 2 cos (2 x) - \frac{1}{2} sin (6 x)

Add a constant to the solution

= - 2 cos (2 x) - \frac{1}{2} sin (6 x) + C

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What is the integral of 4sin(2x)-3cos(6x) ?
The integral of 4sin(2x)-3cos(6x) is -2cos(2x)-1/2 sin(6x)+C