What is the integral of cos^3(x/9) ?

The integral of cos^3(x/9) is 9(sin(x/9)-(sin^3(x/9))/3)+C

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integral of cos^3(x/9)

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

9 (sin (\frac{x}{9}) - \frac{sin ^{3} ( \frac{x}{9} )}{3}) + C

Solution steps

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

Apply u-substitution

: \int cos^{3} (u) \cdot 9 d u

= \int cos^{3} (u) \cdot 9 d u

Take the constant out:

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

= 9 \cdot \int cos^{3} (u) d u

Rewrite using trig identities

= 9 \cdot \int (1 - sin^{2} (u)) cos (u) d u

Apply u-substitution

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

= 9 \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

= 9 (\int 1 d v - \int v^{2} d v)

\int 1 d v = v

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

= 9 (v - \frac{v ^{3}}{3})

Substitute back

= 9 (sin (\frac{x}{9}) - \frac{sin ^{3} ( \frac{x}{9} )}{3})

Add a constant to the solution

= 9 (sin (\frac{x}{9}) - \frac{sin ^{3} ( \frac{x}{9} )}{3}) + C

\int sin^{4} (z)

f (x) = 4 x - 3 x^{2}, (2, - 4)

y = \frac{x}{1 - 5 x}

(x^{2} + 1) y^{^{'}} + 3 x (y - 1) = 0

\frac{d}{d x} ((- 25 x^{2} + 9)^{2})

What is the integral of cos^3(x/9) ?
The integral of cos^3(x/9) is 9(sin(x/9)-(sin^3(x/9))/3)+C