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

f (x) = x^{4} + 5 x - 2, (1, 4)

\int \frac{e ^{4 r}}{r} d r

x \to - 3 - lim (\frac{x + 4}{x + 3})

\int \frac{1}{7 - 6 x} d x

x \to 5 - lim (\frac{x + 1}{x - 5})

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