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

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

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integral of 1-cos(6x)

integral

1 - cos (6 x)

x - \frac{1}{6} sin (6 x) + C

Solution steps

\int 1 - cos (6 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 cos (6 x) d x

\int 1 d x = x

\int cos (6 x) d x = \frac{1}{6} sin (6 x)

= x - \frac{1}{6} sin (6 x)

Add a constant to the solution

= x - \frac{1}{6} sin (6 x) + C

(5 x - 4)^{- 3}

∣ x ∣^{2}

sin (y^{3})

\frac{( e ^{7} x )}{( 2 + e ^{7} x )}

x^{2} - 8 x + 7

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