What is the integral of cos^3(3x)sin^{1/3}(3x) ?

The integral of cos^3(3x)sin^{1/3}(3x) is 1/4 sin^{4/3}(3x)-1/10 sin^{10/3}(3x)+C

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integral of cos^3(3x)sin^{1/3}(3x)

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

\frac{1}{4} sin^{\frac{4}{3}} (3 x) - \frac{1}{10} sin^{\frac{10}{3}} (3 x) + C

Solution steps

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

Apply u-substitution

: \int cos^{3} (u) 3 sin (u) \frac{1}{3} d u

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

Take the constant out:

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

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

Rewrite using trig identities

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

Apply u-substitution

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

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

Expand

3 v (1 - v^{2}) : 3 v - v^{\frac{7}{3}}

= \frac{1}{3} \cdot \int 3 v - v^{\frac{7}{3}} 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} (\int 3 v d v - \int v^{\frac{7}{3}} d v)

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

\int v^{\frac{7}{3}} d v = \frac{3}{10} v^{\frac{10}{3}}

= \frac{1}{3} (\frac{3}{4} v^{\frac{4}{3}} - \frac{3}{10} v^{\frac{10}{3}})

Substitute back

= \frac{1}{3} (\frac{3}{4} sin^{\frac{4}{3}} (3 x) - \frac{3}{10} sin^{\frac{10}{3}} (3 x))

Simplify

\frac{1}{3} (\frac{3}{4} sin^{\frac{4}{3}} (3 x) - \frac{3}{10} sin^{\frac{10}{3}} (3 x)) : \frac{1}{4} sin^{\frac{4}{3}} (3 x) - \frac{1}{10} sin^{\frac{10}{3}} (3 x)

= \frac{1}{4} sin^{\frac{4}{3}} (3 x) - \frac{1}{10} sin^{\frac{10}{3}} (3 x)

Add a constant to the solution

= \frac{1}{4} sin^{\frac{4}{3}} (3 x) - \frac{1}{10} sin^{\frac{10}{3}} (3 x) + C

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What is the integral of cos^3(3x)sin^{1/3}(3x) ?
The integral of cos^3(3x)sin^{1/3}(3x) is 1/4 sin^{4/3}(3x)-1/10 sin^{10/3}(3x)+C