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

The integral of 1/2 (1-cos(x))^2 is 1/8 (6x-8sin(x)+sin(2x))+C

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

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

\frac{1}{8} (6 x - 8 sin (x) + sin (2 x)) + C

Solution steps

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

Expand

(1 - cos (x))^{2} : 1 - 2 cos (x) + cos^{2} (x)

= \frac{1}{2} \cdot \int 1 - 2 cos (x) + 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 2 cos (x) d x + \int cos^{2} (x) d x)

\int 1 d x = x

\int 2 cos (x) d x = 2 sin (x)

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

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

Simplify

\frac{1}{2} (x - 2 sin (x) + \frac{1}{2} (x + \frac{1}{2} sin (2 x))) : \frac{1}{8} (6 x - 8 sin (x) + sin (2 x))

= \frac{1}{8} (6 x - 8 sin (x) + sin (2 x))

Add a constant to the solution

= \frac{1}{8} (6 x - 8 sin (x) + sin (2 x)) + C

\int_{0}^{1} x 4 + 5 x d x

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

y^{^{'}} = - y^{3}, y (0) = a

1 + x^{2}

\frac{d}{d x} (\frac{e ^{x}}{9 x ^{2}})

What is the integral of 1/2 (1-cos(x))^2 ?
The integral of 1/2 (1-cos(x))^2 is 1/8 (6x-8sin(x)+sin(2x))+C