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)

\int 1 - cos (6 x) d 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

t \to 1 lim (\frac{\frac{1}{t} - 1}{t - 1})

n = 0 \sum \infty n^{2}

\frac{d}{d x} (ln (x^{2} - y))

f (x) = (4 x + 5) (5 x - 6)

(0.5) (1.6)

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