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

The integral of cos^3(7x) is 1/7 (sin(7x)-(sin^3(7x))/3)+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(7x)

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

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

Solution steps

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

Apply u-substitution

: \int cos^{3} (u) \frac{1}{7} d u

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

Take the constant out:

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

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

Rewrite using trig identities

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

Apply u-substitution

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

= \frac{1}{7} \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

= \frac{1}{7} (\int 1 d v - \int v^{2} d v)

\int 1 d v = v

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

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

Substitute back

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

Add a constant to the solution

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

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

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

\int \frac{1}{5 t ^{4}} d t

x sin (x^{2})

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

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