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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

Rewrite using trig identities

= \frac{1}{3} \cdot \int (1 - sin^{2} (u)) cos (u) sin^{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{sin ^{5} ( 3 x )}{5} - \frac{sin ^{7} ( 3 x )}{7})

Add a constant to the solution

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

x \to - \infty lim (6 - 2^{x})

\int_{- 1}^{2} (2 + x - x^{2})^{2} d x

\frac{\partial}{\partial y} (x^{2} - 2 x y)

x \to - 3 lim (\frac{x - 7}{x + 3})

1 + (x at (4.3))

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