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

\int \frac{x}{( x ^{2} + 6 x + 13 )} d x

x \to 2 lim (e^{2})

y^{^{'''}} - y^{^{''}} - 20 y^{^{'}} = 0

\int \frac{1}{( 1 - x ^{2} )} d x

2 x^{3} + 36 x^{2} + 210 x + 14

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