What is the integral of 1/3 sin^3(kx) ?

The integral of 1/3 sin^3(kx) is 1/(3k)(-cos(kx)+1/3 cos^3(kx))+C

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integral of 1/3 sin^3(kx)

\int \frac{1}{3} sin^{3} (k x) d x

\frac{1}{3 k} (- cos (k x) + \frac{1}{3} cos^{3} (k x)) + C

Solution steps

\int \frac{1}{3} sin^{3} (k x) d x

Take the constant out:

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

= \frac{1}{3} \cdot \int sin^{3} (k x) d x

Apply u-substitution

: \int \frac{sin ^{3} ( u )}{k} d u

= \frac{1}{3} \cdot \int \frac{sin ^{3} ( u )}{k} d u

Take the constant out:

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

= \frac{1}{3} \cdot \frac{1}{k} \cdot \int sin^{3} (u) d u

Rewrite using trig identities

= \frac{1}{3} \cdot \frac{1}{k} \cdot \int (1 - cos^{2} (u)) sin (u) d u

Apply u-substitution

: \int - 1 + v^{2} d v

= \frac{1}{3} \cdot \frac{1}{k} \cdot \int - 1 + v^{2} d v

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}{3} \cdot \frac{1}{k} (- \int 1 d v + \int v^{2} d v)

\int 1 d v = v

\int v^{2} d v = \frac{v ^{3}}{3}

= \frac{1}{3} \cdot \frac{1}{k} (- v + \frac{v ^{3}}{3})

Substitute back

= \frac{1}{3} \cdot \frac{1}{k} (- cos (k x) + \frac{cos ^{3} ( k x )}{3})

Simplify

\frac{1}{3} \cdot \frac{1}{k} (- cos (k x) + \frac{cos ^{3} ( k x )}{3}) : \frac{1}{3 k} (- cos (k x) + \frac{1}{3} cos^{3} (k x))

= \frac{1}{3 k} (- cos (k x) + \frac{1}{3} cos^{3} (k x))

Add a constant to the solution

= \frac{1}{3 k} (- cos (k x) + \frac{1}{3} cos^{3} (k x)) + C

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What is the integral of 1/3 sin^3(kx) ?
The integral of 1/3 sin^3(kx) is 1/(3k)(-cos(kx)+1/3 cos^3(kx))+C