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

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

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

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

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

Solution steps

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

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

\int 3 cos (4 x) d x = \frac{3}{4} sin (4 x)

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

Simplify

= 2 cos (4 x) + \frac{3}{4} sin (4 x)

Add a constant to the solution

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

\int \frac{1}{2 x ^{2} + 2 x} d x

a re a 4 x, \frac{4 x}{3}

\frac{d x}{d t} = \frac{4 t}{x}

d er i v a t i v e f (x) = \frac{ln ( x )}{38 + x}

\int \frac{5 x ^{2} + 20 x + 6}{x ^{3} + 2 x ^{2} + x} d x

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