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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 3cos^3(2x)

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

\frac{3}{2} sin (2 x) - \frac{1}{2} sin^{3} (2 x) + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

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

Take the constant out:

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

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

Rewrite using trig identities

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

Apply u-substitution

: \int 1 - v^{2} d v

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

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

\int 1 d v = v

\int v^{2} d v = \frac{v ^{3}}{3}

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

Substitute back

= 3 \cdot \frac{1}{2} (sin (2 x) - \frac{sin ^{3} ( 2 x )}{3})

Simplify

3 \cdot \frac{1}{2} (sin (2 x) - \frac{sin ^{3} ( 2 x )}{3}) : \frac{3}{2} sin (2 x) - \frac{1}{2} sin^{3} (2 x)

= \frac{3}{2} sin (2 x) - \frac{1}{2} sin^{3} (2 x)

Add a constant to the solution

= \frac{3}{2} sin (2 x) - \frac{1}{2} sin^{3} (2 x) + C

e^{x} cos (x)

\frac{d}{d t} (2)

f (x) = (- 9 x^{3} + 2 x^{2})^{5}

\int e^{x} \cdot 2 d x

\frac{x}{x + 1}

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