What is the integral of (1-cos(4x))/2 ?

The integral of (1-cos(4x))/2 is 1/2 (x-1/4 sin(4x))+C

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integral of (1-cos(4x))/2

\int \frac{1 - cos ( 4 x )}{2} d x

\frac{1}{2} (x - \frac{1}{4} sin (4 x)) + C

Solution steps

\int \frac{1 - cos ( 4 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 (4 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 (4 x) d x)

\int 1 d x = x

\int cos (4 x) d x = \frac{1}{4} sin (4 x)

= \frac{1}{2} (x - \frac{1}{4} sin (4 x))

Add a constant to the solution

= \frac{1}{2} (x - \frac{1}{4} sin (4 x)) + C

\frac{d}{d x} (7 + 7 x)

\frac{\partial}{\partial y} (4 x - 4 y)

\frac{d}{d x} (\frac{4 x ^{3}}{3} + 2 x + cos (x) + 1)

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

y = \frac{2}{x - 2}, (6, \frac{2}{5})

What is the integral of (1-cos(4x))/2 ?
The integral of (1-cos(4x))/2 is 1/2 (x-1/4 sin(4x))+C