What is the integral of sin^3(18x) ?

The integral of sin^3(18x) is 1/18 (-cos(18x)+(cos^3(18x))/3)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of sin^3(18x)

\int sin^{3} (18 x) d x

\frac{1}{18} (- cos (18 x) + \frac{cos ^{3} ( 18 x )}{3}) + C

Solution steps

\int sin^{3} (18 x) d x

Apply u-substitution

: \int sin^{3} (u) \frac{1}{18} d u

= \int sin^{3} (u) \frac{1}{18} d u

Take the constant out:

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

= \frac{1}{18} \cdot \int sin^{3} (u) d u

Rewrite using trig identities

= \frac{1}{18} \cdot \int (1 - cos^{2} (u)) sin (u) d u

Apply u-substitution

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

= \frac{1}{18} \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}{18} (- \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}{18} (- v + \frac{v ^{3}}{3})

Substitute back

= \frac{1}{18} (- cos (18 x) + \frac{cos ^{3} ( 18 x )}{3})

Add a constant to the solution

= \frac{1}{18} (- cos (18 x) + \frac{cos ^{3} ( 18 x )}{3}) + C

\int sec^{4 x} (d) x d x

\frac{d}{d x} (\frac{x ^{2} - 64}{x + 8})

g (t) = (9 t + 5)^{2} (5 t - 6)^{- 3}

(3 x + 2) + (3 y - 3) y^{^{'}} = 0

\int y (y + 1) e^{- 2 y} d y

What is the integral of sin^3(18x) ?
The integral of sin^3(18x) is 1/18 (-cos(18x)+(cos^3(18x))/3)+C