What is the integral of sin^3(3x)cos^2(3x) ?

The integral of sin^3(3x)cos^2(3x) is 1/3 (-(cos^3(3x))/3+(cos^5(3x))/5)+C

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integral of sin^3(3x)cos^2(3x)

\int sin^{3} (3 x) cos^{2} (3 x) d x

\frac{1}{3} (- \frac{cos ^{3} ( 3 x )}{3} + \frac{cos ^{5} ( 3 x )}{5}) + C

Solution steps

\int sin^{3} (3 x) cos^{2} (3 x) d x

Apply u-substitution

: \int sin^{3} (u) cos^{2} (u) \frac{1}{3} d u

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

Rewrite using trig identities

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

Apply u-substitution

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

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

Expand

- v^{2} (1 - v^{2}) : - v^{2} + v^{4}

= \frac{1}{3} \cdot \int - v^{2} + v^{4} 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^{2} d v + \int v^{4} d v)

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

\int v^{4} d v = \frac{v ^{5}}{5}

= \frac{1}{3} (- \frac{v ^{3}}{3} + \frac{v ^{5}}{5})

Substitute back

= \frac{1}{3} (- \frac{cos ^{3} ( 3 x )}{3} + \frac{cos ^{5} ( 3 x )}{5})

Add a constant to the solution

= \frac{1}{3} (- \frac{cos ^{3} ( 3 x )}{3} + \frac{cos ^{5} ( 3 x )}{5}) + C

f (x) = \frac{6 x}{x + 3}, (- 1, - 3)

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

\frac{\partial}{\partial z} (x^{2} z^{2} sin (4 y))

\frac{\partial}{\partial x} (sin (\frac{2 x}{y}))

- 5 w^{- 7} + 3 w

What is the integral of sin^3(3x)cos^2(3x) ?
The integral of sin^3(3x)cos^2(3x) is 1/3 (-(cos^3(3x))/3+(cos^5(3x))/5)+C