What is the integral of (2sin(4x)-8cos(3x)) ?

The integral of (2sin(4x)-8cos(3x)) is -1/2 cos(4x)-8/3 sin(3x)+C

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integral of (2sin(4x)-8cos(3x))

\int (2 sin (4 x) - 8 cos (3 x)) d x

- \frac{1}{2} cos (4 x) - \frac{8}{3} sin (3 x) + C

Solution steps

\int 2 sin (4 x) - 8 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 2 sin (4 x) d x - \int 8 cos (3 x) d x

\int 2 sin (4 x) d x = - \frac{1}{2} cos (4 x)

\int 8 cos (3 x) d x = \frac{8}{3} sin (3 x)

= - \frac{1}{2} cos (4 x) - \frac{8}{3} sin (3 x)

Add a constant to the solution

= - \frac{1}{2} cos (4 x) - \frac{8}{3} sin (3 x) + C

\int_{0}^{1} x e^{- 5 x} d x

\frac{d}{d x} ((x + 4)^{3} (x - 3)^{6})

x \to 4 lim (2 - x)

y = 3 x - 7 x

x + 4 y + 2 x y

What is the integral of (2sin(4x)-8cos(3x)) ?
The integral of (2sin(4x)-8cos(3x)) is -1/2 cos(4x)-8/3 sin(3x)+C