What is the integral of 5cos^3(4x) ?

The integral of 5cos^3(4x) is 5/4 (sin(4x)-1/3 sin^3(4x))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 5cos^3(4x)

\int 5 cos^{3} (4 x) d x

\frac{5}{4} (sin (4 x) - \frac{1}{3} sin^{3} (4 x)) + C

Solution steps

\int 5 cos^{3} (4 x) d x

Take the constant out:

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

= 5 \cdot \int cos^{3} (4 x) d x

Apply u-substitution

: \int cos^{3} (u) \frac{1}{4} d u

= 5 \cdot \int cos^{3} (u) \frac{1}{4} d u

Take the constant out:

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

= 5 \cdot \frac{1}{4} \cdot \int cos^{3} (u) d u

Rewrite using trig identities

= 5 \cdot \frac{1}{4} \cdot \int (1 - sin^{2} (u)) cos (u) d u

Apply u-substitution

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

= 5 \cdot \frac{1}{4} \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

= 5 \cdot \frac{1}{4} (\int 1 d v - \int v^{2} d v)

\int 1 d v = v

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

= 5 \cdot \frac{1}{4} (v - \frac{v ^{3}}{3})

Substitute back

= 5 \cdot \frac{1}{4} (sin (4 x) - \frac{sin ^{3} ( 4 x )}{3})

Simplify

5 \cdot \frac{1}{4} (sin (4 x) - \frac{sin ^{3} ( 4 x )}{3}) : \frac{5}{4} (sin (4 x) - \frac{1}{3} sin^{3} (4 x))

= \frac{5}{4} (sin (4 x) - \frac{1}{3} sin^{3} (4 x))

Add a constant to the solution

= \frac{5}{4} (sin (4 x) - \frac{1}{3} sin^{3} (4 x)) + C

\frac{d}{d x} (x^{\frac{2}{3}} + 1)

\int - t cos (t) d t

f (x) = \frac{1}{x ^{2}}, (1, 1)

\frac{d}{d x} (\frac{10}{x ^{2} - 4 x + 1})

\int (x + 7) (x - 3) d x

What is the integral of 5cos^3(4x) ?
The integral of 5cos^3(4x) is 5/4 (sin(4x)-1/3 sin^3(4x))+C