What is the integral of 7(cos(4x))^3 ?

The integral of 7(cos(4x))^3 is 7/4 (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 7(cos(4x))^3

\int 7 (cos (4 x))^{3} d x

\frac{7}{4} (sin (4 x) - \frac{1}{3} sin^{3} (4 x)) + C

Solution steps

\int 7 (cos (4 x))^{3} d x

Take the constant out:

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

= 7 \cdot \int (cos (4 x))^{3} d x

Apply u-substitution

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

= 7 \cdot \int \frac{1}{4} cos^{3} (u) d u

Take the constant out:

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

= 7 \cdot \frac{1}{4} \cdot \int cos^{3} (u) d u

Rewrite using trig identities

= 7 \cdot \frac{1}{4} \cdot \int (1 - sin^{2} (u)) cos (u) d u

Apply u-substitution

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

= 7 \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

= 7 \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}

= 7 \cdot \frac{1}{4} (v - \frac{v ^{3}}{3})

Substitute back

= 7 \cdot \frac{1}{4} (sin (4 x) - \frac{sin ^{3} ( 4 x )}{3})

Simplify

7 \cdot \frac{1}{4} (sin (4 x) - \frac{sin ^{3} ( 4 x )}{3}) : \frac{7}{4} (sin (4 x) - \frac{1}{3} sin^{3} (4 x))

= \frac{7}{4} (sin (4 x) - \frac{1}{3} sin^{3} (4 x))

Add a constant to the solution

= \frac{7}{4} (sin (4 x) - \frac{1}{3} sin^{3} (4 x)) + C

\int \frac{11 x ^{2} + 5 x - 4}{2 x ^{3} - x ^{2}} d x

\int x^{2} x^{3} - 5 d x

\int \frac{x + 1}{x ^{4} + 3 x ^{2}} d x

\int \frac{x ^{2}}{x ^{2} + 4 x + 5} d x

\int \frac{4 x ^{2} + 9 x - 1}{x ^{3} + 2 x ^{2} - x - 2} d x

What is the integral of 7(cos(4x))^3 ?
The integral of 7(cos(4x))^3 is 7/4 (sin(4x)-1/3 sin^3(4x))+C