What is the integral of 8sin(2x)-3cos(1/5 x) ?

The integral of 8sin(2x)-3cos(1/5 x) is -4cos(2x)-15sin(1/5 x)+C

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integral of 8sin(2x)-3cos(1/5 x)

\int 8 sin (2 x) - 3 cos (\frac{1}{5} x) d x

- 4 cos (2 x) - 15 sin (\frac{1}{5} x) + C

Solution steps

\int 8 sin (2 x) - 3 cos (\frac{1}{5} 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 (2 x) d x - \int 3 cos (\frac{1}{5} x) d x

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

\int 3 cos (\frac{1}{5} x) d x = 15 sin (\frac{1}{5} x)

= - 4 cos (2 x) - 15 sin (\frac{1}{5} x)

Add a constant to the solution

= - 4 cos (2 x) - 15 sin (\frac{1}{5} x) + C

\int \frac{x}{( x + 2 ) ^{4}} d x

\int \frac{5 x - 13}{( x - 3 ) ( x - 2 )} d x

\int (x - sin (x) + cos (x)) d x

\int \frac{1}{( a - x )} d x

What is the integral of 8sin(2x)-3cos(1/5 x) ?
The integral of 8sin(2x)-3cos(1/5 x) is -4cos(2x)-15sin(1/5 x)+C