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

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

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integral of (1-cos(2x))/2

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

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

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \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

= \frac{1}{2} (\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)

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

Add a constant to the solution

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

\int + x^{2} sec^{2} (x^{3} + 1) d x

\frac{\partial}{\partial x} (x^{3} (1 + y) + z^{2})

\frac{d}{d x} (\frac{x ^{2}}{( 5 + x )})

arctan (x) - \frac{π}{4}

\frac{\partial}{\partial x} (\frac{x ^{2}}{1 + y ^{2}})

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