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

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

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integral of 2sin(3x)+3sin(2x)

\int 2 sin (3 x) + 3 sin (2 x) d x

- \frac{2}{3} cos (3 x) - \frac{3}{2} cos (2 x) + C

Solution steps

\int 2 sin (3 x) + 3 sin (2 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 2 sin (3 x) d x + \int 3 sin (2 x) d x

\int 2 sin (3 x) d x = - \frac{2}{3} cos (3 x)

\int 3 sin (2 x) d x = - \frac{3}{2} cos (2 x)

= - \frac{2}{3} cos (3 x) - \frac{3}{2} cos (2 x)

Add a constant to the solution

= - \frac{2}{3} cos (3 x) - \frac{3}{2} cos (2 x) + C

\frac{d}{d x} (ln (cosh (4 x)))

\int tan^{2} (2 y) d y

\frac{d}{d x} ((x + 5)^{2})

x \to \infty lim (3 x - 1)

t \to 0 lim (\frac{sin ( 3 t )}{t} + 2)

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