What is the integral of sin^3(6x)cos^2(6x) ?

The integral of sin^3(6x)cos^2(6x) is 1/6 (-(cos^3(6x))/3+(cos^5(6x))/5)+C

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integral of sin^3(6x)cos^2(6x)

\int sin^{3} (6 x) cos^{2} (6 x) d x

\frac{1}{6} (- \frac{cos ^{3} ( 6 x )}{3} + \frac{cos ^{5} ( 6 x )}{5}) + C

Solution steps

\int sin^{3} (6 x) cos^{2} (6 x) d x

Apply u-substitution

: \int sin^{3} (u) cos^{2} (u) \frac{1}{6} d u

= \int sin^{3} (u) cos^{2} (u) \frac{1}{6} d u

Take the constant out:

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

= \frac{1}{6} \cdot \int sin^{3} (u) cos^{2} (u) d u

Rewrite using trig identities

= \frac{1}{6} \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}{6} \cdot \int - v^{2} (1 - v^{2}) d v

Expand

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

= \frac{1}{6} \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}{6} (- \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}{6} (- \frac{v ^{3}}{3} + \frac{v ^{5}}{5})

Substitute back

= \frac{1}{6} (- \frac{cos ^{3} ( 6 x )}{3} + \frac{cos ^{5} ( 6 x )}{5})

Add a constant to the solution

= \frac{1}{6} (- \frac{cos ^{3} ( 6 x )}{3} + \frac{cos ^{5} ( 6 x )}{5}) + C

\int_{0}^{3} π (\frac{1}{x + 5})^{2} d x

y = 3 x + 1, y = x + 1

\int 5 u^{5} u^{6} + 5 d u

\frac{d}{d x} (\frac{π}{( 2 x ) ^{4}})

\frac{d y}{d x} = 2 x y - 6 x

What is the integral of sin^3(6x)cos^2(6x) ?
The integral of sin^3(6x)cos^2(6x) is 1/6 (-(cos^3(6x))/3+(cos^5(6x))/5)+C