What is the integral of (2sin(kr))/k ?

The integral of (2sin(kr))/k is -2/(k^2)cos(kr)+C

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integral of (2sin(kr))/k

\int \frac{2 sin ( k r )}{k} d r

- \frac{2}{k ^{2}} cos (k r) + C

Solution steps

\int \frac{2 sin ( k r )}{k} d r

Take the constant out:

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

= \frac{2}{k} \cdot \int sin (k r) d r

Apply u-substitution

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

Take the constant out:

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

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

Use the common integral:

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

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

Substitute back

u = k r

= \frac{2}{k} \cdot \frac{1}{k} (- cos (k r))

Simplify

\frac{2}{k} \cdot \frac{1}{k} (- cos (k r)) : - \frac{2}{k ^{2}} cos (k r)

= - \frac{2}{k ^{2}} cos (k r)

Add a constant to the solution

= - \frac{2}{k ^{2}} cos (k r) + C

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