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)

Add a constant to the solution

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

(2 + x)^{- 2}

x \to \infty lim (\frac{2 ( x - 8 ) ^{35}}{( x - 1 ) ^{35}})

f (x) = \frac{x + 1}{x ^{2} + 1}

\frac{d y}{d t} = y^{3}, y (0) = 1

\frac{d}{d x} (\frac{432}{x})

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