What is the integral of 1-6sin(3x) ?

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

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integral of 1-6sin(3x)

\int 1 - 6 sin (3 x) d x

x + 2 cos (3 x) + C

Solution steps

\int 1 - 6 sin (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 1 d x - \int 6 sin (3 x) d x

\int 1 d x = x

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

= x - (- 2 cos (3 x))

Simplify

= x + 2 cos (3 x)

Add a constant to the solution

= x + 2 cos (3 x) + C

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