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

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

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

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

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

Solution steps

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

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

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

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

Simplify

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

Add a constant to the solution

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

x^{2} e^{19 x}

\frac{2 x ^{5} + 6}{x ^{3}}

x^{2}, x + 2, 2, - 1

\int - 5 x^{5} e^{x^{6}} d x

\int 9 x d x

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