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

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

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integral of xcos(1/2 x)

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

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

Solution steps

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

Simplify

\frac{1}{2} x : \frac{x}{2}

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

Apply u-substitution

: \int 4 u cos (u) d u

= \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 \int u cos (u) d u

Apply Integration By Parts

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

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

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

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

Substitute back

u = \frac{x}{2}

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

Simplify

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

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

Add a constant to the solution

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

y^{^{''}} + 3 y^{^{'}} + 2 y = x + 6 e^{x}, y (0) = 0, y^{^{'}} (0) = 1

y = sin (x y) + π, (1, π)

7 \frac{d ^{2} y}{d x ^{2}} + 4 \frac{d y}{d x} + 9 y = 0

f (x) = x^{4} + 3 x^{2} - x, (1, 3)

\int_{0}^{a} x d x

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