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

The integral of cos^3(3x)sin^3(3x) is 1/3 (-(cos^4(3x))/4+(cos^6(3x))/6)+C

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

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

\frac{1}{3} (- \frac{cos ^{4} ( 3 x )}{4} + \frac{cos ^{6} ( 3 x )}{6}) + C

Solution steps

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

Apply u-substitution

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

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

Rewrite using trig identities

= \frac{1}{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

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

Expand

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

= \frac{1}{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

= \frac{1}{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}

= \frac{1}{3} (- \frac{v ^{4}}{4} + \frac{v ^{6}}{6})

Substitute back

= \frac{1}{3} (- \frac{cos ^{4} ( 3 x )}{4} + \frac{cos ^{6} ( 3 x )}{6})

Add a constant to the solution

= \frac{1}{3} (- \frac{cos ^{4} ( 3 x )}{4} + \frac{cos ^{6} ( 3 x )}{6}) + C

f (x) = (e^{x} + x^{π}) (x^{2} + x + 1)

y^{^{'}} = 0.1 y (10 - y)

\frac{\partial}{\partial x} (ln (\frac{a}{x}))

\int_{0}^{5} 5 d x

n = 6 \sum \infty \frac{5}{n ^{2} - 25}

What is the integral of cos^3(3x)sin^3(3x) ?
The integral of cos^3(3x)sin^3(3x) is 1/3 (-(cos^4(3x))/4+(cos^6(3x))/6)+C