What is the integral of sin^3(x/6) ?

The integral of sin^3(x/6) is 6(-cos(x/6)+(cos^3(x/6))/3)+C

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integral of sin^3(x/6)

\int sin^{3} (\frac{x}{6}) d x

6 (- cos (\frac{x}{6}) + \frac{cos ^{3} ( \frac{x}{6} )}{3}) + C

Solution steps

\int sin^{3} (\frac{x}{6}) d x

Apply u-substitution

: \int sin^{3} (u) \cdot 6 d u

= \int sin^{3} (u) \cdot 6 d u

Take the constant out:

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

= 6 \cdot \int sin^{3} (u) d u

Rewrite using trig identities

= 6 \cdot \int (1 - cos^{2} (u)) sin (u) d u

Apply u-substitution

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

= 6 \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

= 6 (- \int 1 d v + \int v^{2} d v)

\int 1 d v = v

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

= 6 (- v + \frac{v ^{3}}{3})

Substitute back

= 6 (- cos (\frac{x}{6}) + \frac{cos ^{3} ( \frac{x}{6} )}{3})

Add a constant to the solution

= 6 (- cos (\frac{x}{6}) + \frac{cos ^{3} ( \frac{x}{6} )}{3}) + C

x (\frac{d y}{d x}) = y^{2} - y

y = 2 x, y = 2 x^{3}

\frac{\partial}{\partial x} (\frac{x}{y - x y})

y = 5 7 e^{- 2 x} + 6

What is the integral of sin^3(x/6) ?
The integral of sin^3(x/6) is 6(-cos(x/6)+(cos^3(x/6))/3)+C