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)

Simplify

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

Add a constant to the solution

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

\frac{d y}{d x} = \frac{15 x}{y}

\frac{d}{d y} (e^{x e^{y}})

x, x - 2, y = 0

x \to 0 lim (\frac{sin ( k x )}{x})

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