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

The integral of (6sin(3x)-cos(2x)) is -2cos(3x)-1/2 sin(2x)+C

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

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

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

Solution steps

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

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

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

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

Add a constant to the solution

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

\int x - 4 d x

x \to 0 + lim (\frac{3}{x ^{2} - x})

\int \frac{16 s + 16}{( s ^{2} + 1 ) ( s - 1 ) ^{3}} d s

x \to 0.999 lim (x^{2} + 2)

\frac{2 s}{s ^{2} + 6}

What is the integral of (6sin(3x)-cos(2x)) ?
The integral of (6sin(3x)-cos(2x)) is -2cos(3x)-1/2 sin(2x)+C