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

The integral of-4*sin(2x)+3*cos(3x) is 2cos(2x)+sin(3x)+C

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integral of-4sin(2x)+3cos(3x)

\int - 4 \cdot sin (2 x) + 3 \cdot 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

\frac{d ^{2} y}{d x ^{2}} = \frac{d y}{d x}

n = 1 \sum \infty \frac{( - 1 ) ^{n - 1}}{n !}

\frac{x ^{2} - 3 x}{2}

ln (x + x^{2} + 1)

x \to \infty lim (\frac{x}{( ln ( x ) ) ^{3} + 2 x})

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