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

The integral of 2sin^3(3x)cos^3(3x) is 2/3 (-1/4 cos^4(3x)+1/6 cos^6(3x))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

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

\frac{2}{3} (- \frac{1}{4} cos^{4} (3 x) + \frac{1}{6} cos^{6} (3 x)) + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

= 2 \cdot \int sin^{3} (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

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

Rewrite using trig identities

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

Apply u-substitution

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

Expand

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

= 2 \cdot \frac{1}{3} \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

= 2 \cdot \frac{1}{3} (- \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}

= 2 \cdot \frac{1}{3} (- \frac{v ^{4}}{4} + \frac{v ^{6}}{6})

Substitute back

= 2 \cdot \frac{1}{3} (- \frac{cos ^{4} ( 3 x )}{4} + \frac{cos ^{6} ( 3 x )}{6})

Simplify

2 \cdot \frac{1}{3} (- \frac{cos ^{4} ( 3 x )}{4} + \frac{cos ^{6} ( 3 x )}{6}) : \frac{2}{3} (- \frac{1}{4} cos^{4} (3 x) + \frac{1}{6} cos^{6} (3 x))

= \frac{2}{3} (- \frac{1}{4} cos^{4} (3 x) + \frac{1}{6} cos^{6} (3 x))

Add a constant to the solution

= \frac{2}{3} (- \frac{1}{4} cos^{4} (3 x) + \frac{1}{6} cos^{6} (3 x)) + C

What is the integral of 2sin^3(3x)cos^3(3x) ?
The integral of 2sin^3(3x)cos^3(3x) is 2/3 (-1/4 cos^4(3x)+1/6 cos^6(3x))+C