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

The integral of sin(2x)-cos(3x) is -1/2 cos(2x)-1/3 sin(3x)+C

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

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

- \frac{1}{2} cos (2 x) - \frac{1}{3} sin (3 x) + C

Solution steps

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

\int sin (2 x) d x = - \frac{1}{2} cos (2 x)

\int cos (3 x) d x = \frac{1}{3} sin (3 x)

= - \frac{1}{2} cos (2 x) - \frac{1}{3} sin (3 x)

Add a constant to the solution

= - \frac{1}{2} cos (2 x) - \frac{1}{3} sin (3 x) + C

y^{^{''}} + y = x^{2}

y^{2} = 4 x, y = 2 x - 4

\frac{d}{d x} (4 x^{2} + 3 x)

\frac{d x}{d t} = \frac{x}{t}

\frac{\partial}{\partial x} (\frac{1}{∣ x - a ∣})

What is the integral of sin(2x)-cos(3x) ?
The integral of sin(2x)-cos(3x) is -1/2 cos(2x)-1/3 sin(3x)+C