What is the integral of 1-cos(2x) ?

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

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integral of 1-cos(2x)

\int 1 - cos (2 x) d x

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

Solution steps

\int 1 - 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 1 d x - \int cos (2 x) d x

\int 1 d x = x

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

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

Add a constant to the solution

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

\frac{1}{s ^{2} + 2 s}

\int u^{6} d u

x \to 5 lim (sin (\frac{π x}{3}))

(3 x^{2} - 2 x y) d x + (2 y - x^{2}) d y = 0

\int_{- 1}^{3} (x^{2} - 2 x) d x

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