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

The integral of-4xcos(x/2) is -16(1/2 xsin(x/2)+cos(x/2))+C

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integral of-4xcos(x/2)

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

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

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

: \int 4 u cos (u) d u

= - 4 \cdot \int 4 u cos (u) d u

Take the constant out:

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

= - 4 \cdot 4 \cdot \int u cos (u) d u

Apply Integration By Parts

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

= - 4 \cdot 4 (u sin (u) - \int sin (u) d u)

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

= - 4 \cdot 4 (u sin (u) - (- cos (u)))

Substitute back

u = \frac{x}{2}

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

Simplify

- 4 \cdot 4 (\frac{x}{2} sin (\frac{x}{2}) - (- cos (\frac{x}{2}))) : - 16 (\frac{1}{2} x sin (\frac{x}{2}) + cos (\frac{x}{2}))

= - 16 (\frac{1}{2} x sin (\frac{x}{2}) + cos (\frac{x}{2}))

Add a constant to the solution

= - 16 (\frac{1}{2} x sin (\frac{x}{2}) + cos (\frac{x}{2})) + C

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What is the integral of-4xcos(x/2) ?
The integral of-4xcos(x/2) is -16(1/2 xsin(x/2)+cos(x/2))+C