What is the integral of sin^3(x/3)cos^3(x/3) ?

The integral of sin^3(x/3)cos^3(x/3) is 3(-(cos^4(x/3))/4+(cos^6(x/3))/6)+C

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integral of sin^3(x/3)cos^3(x/3)

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

3 (- \frac{cos ^{4} ( \frac{x}{3} )}{4} + \frac{cos ^{6} ( \frac{x}{3} )}{6}) + C

Solution steps

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

Apply u-substitution

: \int sin^{3} (u) cos^{3} (u) \cdot 3 d u

= \int sin^{3} (u) cos^{3} (u) \cdot 3 d u

Take the constant out:

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

= 3 \cdot \int sin^{3} (u) cos^{3} (u) d u

Rewrite using trig identities

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

Apply u-substitution

: \int - v^{3} (1 - v^{2}) d v

= 3 \cdot \int - v^{3} (1 - v^{2}) d v

Expand

- v^{3} (1 - v^{2}) : - v^{3} + v^{5}

= 3 \cdot \int - v^{3} + v^{5} 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

= 3 (- \int v^{3} d v + \int v^{5} d v)

\int v^{3} d v = \frac{v ^{4}}{4}

\int v^{5} d v = \frac{v ^{6}}{6}

= 3 (- \frac{v ^{4}}{4} + \frac{v ^{6}}{6})

Substitute back

= 3 (- \frac{cos ^{4} ( \frac{x}{3} )}{4} + \frac{cos ^{6} ( \frac{x}{3} )}{6})

Add a constant to the solution

= 3 (- \frac{cos ^{4} ( \frac{x}{3} )}{4} + \frac{cos ^{6} ( \frac{x}{3} )}{6}) + C

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What is the integral of sin^3(x/3)cos^3(x/3) ?
The integral of sin^3(x/3)cos^3(x/3) is 3(-(cos^4(x/3))/4+(cos^6(x/3))/6)+C