What is the integral of sin^3(3x)cos^4(3x) ?

The integral of sin^3(3x)cos^4(3x) is 1/3 (-(cos^5(3x))/5+(cos^7(3x))/7)+C

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integral of sin^3(3x)cos^4(3x)

\int sin^{3} (3 x) cos^{4} (3 x) d x

\frac{1}{3} (- \frac{cos ^{5} ( 3 x )}{5} + \frac{cos ^{7} ( 3 x )}{7}) + C

Solution steps

\int sin^{3} (3 x) cos^{4} (3 x) d x

Apply u-substitution

: \int sin^{3} (u) cos^{4} (u) \frac{1}{3} d u

= \int sin^{3} (u) cos^{4} (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^{4} (u) d u

Rewrite using trig identities

= \frac{1}{3} \cdot \int (1 - cos^{2} (u)) sin (u) cos^{4} (u) d u

Apply u-substitution

: \int - v^{4} (1 - v^{2}) d v

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

Expand

- v^{4} (1 - v^{2}) : - v^{4} + v^{6}

= \frac{1}{3} \cdot \int - v^{4} + v^{6} 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^{4} d v + \int v^{6} d v)

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

\int v^{6} d v = \frac{v ^{7}}{7}

= \frac{1}{3} (- \frac{v ^{5}}{5} + \frac{v ^{7}}{7})

Substitute back

= \frac{1}{3} (- \frac{cos ^{5} ( 3 x )}{5} + \frac{cos ^{7} ( 3 x )}{7})

Add a constant to the solution

= \frac{1}{3} (- \frac{cos ^{5} ( 3 x )}{5} + \frac{cos ^{7} ( 3 x )}{7}) + C

\frac{\partial}{\partial x} ((1 - x) (1 - y))

x \to 4 + lim (\frac{x - 4}{x ^{2} - 1})

f (x) = 3 x^{6} + 9 x^{2} - 7

\frac{d}{d x} (arccos (- \frac{x}{5}))

(sin (x) - y sin (x) - 2 cos (x)) + cos (x) y^{^{'}} = 0, y (0) = - 1

What is the integral of sin^3(3x)cos^4(3x) ?
The integral of sin^3(3x)cos^4(3x) is 1/3 (-(cos^5(3x))/5+(cos^7(3x))/7)+C