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 \to 0 + lim (x sin (\frac{1}{x}))

P (t) = 0.45 t^{2} - 12 t + 105

\frac{d}{d x} ((cos (x)) sin (x))

\int (\frac{1}{x} - 5 x) d x

t \frac{d y}{d t} + (t + 1) y = t

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