What is the integral of 6*sin(2x)-3*cos(3x) ?

The integral of 6*sin(2x)-3*cos(3x) is -3cos(2x)-sin(3x)+C

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integral of 6sin(2x)-3cos(3x)

\int 6 \cdot sin (2 x) - 3 \cdot cos (3 x) d x

- 3 cos (2 x) - sin (3 x) + C

Solution steps

\int 6 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 6 sin (2 x) d x - \int 3 cos (3 x) d x

\int 6 sin (2 x) d x = - 3 cos (2 x)

\int 3 cos (3 x) d x = sin (3 x)

= - 3 cos (2 x) - sin (3 x)

Add a constant to the solution

= - 3 cos (2 x) - sin (3 x) + C

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

x d y = y + x (cos (\frac{y}{x}))^{2} d x

2 x^{3} + 3 x^{2} - 12 x + 1

x \to 0 + lim (x - \frac{1}{x} + 3)

\int \frac{x ^{2} - 3}{( x + 1 ) ^{3}} d x

What is the integral of 6*sin(2x)-3*cos(3x) ?
The integral of 6*sin(2x)-3*cos(3x) is -3cos(2x)-sin(3x)+C