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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Solution steps

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

Apply u-substitution

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

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

Rewrite using trig identities

= \frac{1}{6} \cdot \int (1 - sin^{2} (u)) cos (u) sin^{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{sin ^{3} ( 6 x )}{3} - \frac{sin ^{5} ( 6 x )}{5})

Add a constant to the solution

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

\int \frac{3 e ^{2 x}}{e ^{2 x} + 11 e ^{x} + 28} d x

\frac{d y}{d x} = a y - b y^{2}

8 sin (4 t) - 5 cos (4 t)

\int_{- 2}^{2} 8 1^{\frac{x}{2}} d x

\frac{( x + 3 )}{( x - 3 )}

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