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

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

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

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

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

Solution steps

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

\int x d x = \frac{x ^{2}}{2}

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

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

Simplify

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

Add a constant to the solution

= \frac{1}{2} (\frac{x ^{2}}{2} + \frac{1}{4} cos (2 x)) + C

f (x) = \frac{2}{3 x ^{2}}

n = 1 \sum \infty 2^{- 2 n}

y^{^{''}} + 4 y^{^{'}} + 4 y = 2 x + 6

y^{^{''}} - 2 y^{^{'}} + 2 y = e^{x} sin (x)

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