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

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

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

\int (- 6 sin (2 x) + 3 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)

Simplify

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

Add a constant to the solution

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

\int_{- 1}^{1} 1 + 4 x^{2} d x

\int (x + \frac{3}{x}) d x

x \to - 1 lim ((1 + x)^{- 1})

\frac{\partial}{\partial x} (\frac{3 x - y}{( x + y ) ^{3}})

y^{^{'}} = \frac{t ^{2} + t y + y ^{2}}{t ^{2}}

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