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

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

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integral of 2x-sin(2x)

\int 2 x - sin (2 x) d x

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

Solution steps

\int 2 x - sin (2 x) d x

Apply u-substitution

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

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

Take the constant out:

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

= \frac{1}{2} \cdot \int u - sin (u) d u

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 u d u - \int sin (u) d u)

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

\int sin (u) d u = - cos (u)

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

Substitute back

u = 2 x

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

Simplify

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

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

Add a constant to the solution

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

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