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

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

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

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

- 3 cos (2 x) - sin (3 x) + C

Solution steps

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

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

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

= - 3 cos (2 x) - sin (3 x)

Add a constant to the solution

= - 3 cos (2 x) - sin (3 x) + C

x \to 1 lim ((x + 2)^{4} (4 x^{2}))

\frac{d}{d x} (\frac{3 csc ( x )}{4 ^{x}})

\int \frac{5}{2} x d x

\frac{\partial}{\partial x} (sec (17 x))

\frac{\partial}{\partial x} (25 - (x^{2} + 3 y^{2}))

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