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

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

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integral of cos^3(x/6)

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

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

Solution steps

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

Apply u-substitution

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

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

Rewrite using trig identities

= 6 \cdot \int (1 - sin^{2} (u)) cos (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 (sin (\frac{x}{6}) - \frac{sin ^{3} ( \frac{x}{6} )}{3})

Add a constant to the solution

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

y^{^{''}} + 7 y^{^{'}} + 6 y = 0, y (0) = 1, y^{^{'}} (0) = 0

\int (\frac{9}{x} - \frac{2}{x ^{3}}) d x

\frac{d}{d x} (x e^{y^{2}})

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

x \to \frac{c}{2 ^{+}} lim ([2 x - 4])

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