What is the integral of sin^3(18x)cos^3(18x) ?

The integral of sin^3(18x)cos^3(18x) is 1/18 (-(cos^4(18x))/4+(cos^6(18x))/6)+C

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integral of sin^3(18x)cos^3(18x)

\int sin^{3} (18 x) cos^{3} (18 x) d x

\frac{1}{18} (- \frac{cos ^{4} ( 18 x )}{4} + \frac{cos ^{6} ( 18 x )}{6}) + C

Solution steps

\int sin^{3} (18 x) cos^{3} (18 x) d x

Apply u-substitution

: \int sin^{3} (u) cos^{3} (u) \frac{1}{18} d u

= \int sin^{3} (u) cos^{3} (u) \frac{1}{18} d u

Take the constant out:

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

= \frac{1}{18} \cdot \int sin^{3} (u) cos^{3} (u) d u

Rewrite using trig identities

= \frac{1}{18} \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}{18} \cdot \int - v^{3} (1 - v^{2}) d v

Expand

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

= \frac{1}{18} \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}{18} (- \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}{18} (- \frac{v ^{4}}{4} + \frac{v ^{6}}{6})

Substitute back

= \frac{1}{18} (- \frac{cos ^{4} ( 18 x )}{4} + \frac{cos ^{6} ( 18 x )}{6})

Add a constant to the solution

= \frac{1}{18} (- \frac{cos ^{4} ( 18 x )}{4} + \frac{cos ^{6} ( 18 x )}{6}) + C

\int \frac{8 x}{( 1 - x ^{2} ) ^{4}} d x

\frac{\partial}{\partial z} (\frac{4 x - 4 z}{y + 4 z})

y y^{^{'}} = x - 2 x^{3}

\frac{d}{d x} (a^{7} + (cos (x))^{7})

x = y^{2} + 4 y - 22, x = 3 y + 8

What is the integral of sin^3(18x)cos^3(18x) ?
The integral of sin^3(18x)cos^3(18x) is 1/18 (-(cos^4(18x))/4+(cos^6(18x))/6)+C