What is the integral of sin^3(ax) ?

The integral of sin^3(ax) is 1/a (-cos(ax)+(cos^3(ax))/3)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of sin^3(ax)

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

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

Solution steps

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

Apply u-substitution

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

Take the constant out:

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

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

Rewrite using trig identities

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

Apply u-substitution

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

Substitute back

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

Add a constant to the solution

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

a re a sin (x), sin (2 x), [0, \frac{π}{3}]

d er i v a t i v e f (x) = \frac{1}{x} - 2 x

\frac{d}{d x} (\frac{3}{4} x^{2} - 5 x^{3})

\int \frac{1}{- 2 x} d x

l a pl a ce t r an s f or m cos^{2} (25 t)

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