What is the integral of cos^3(x/8) ?

The integral of cos^3(x/8) is 8(sin(x/8)-(sin^3(x/8))/3)+C

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integral of cos^3(x/8)

\int cos^{3} (\frac{x}{8}) d x

8 (sin (\frac{x}{8}) - \frac{sin ^{3} ( \frac{x}{8} )}{3}) + C

Solution steps

\int cos^{3} (\frac{x}{8}) d x

Apply u-substitution

: \int cos^{3} (u) \cdot 8 d u

= \int cos^{3} (u) \cdot 8 d u

Take the constant out:

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

= 8 \cdot \int cos^{3} (u) d u

Rewrite using trig identities

= 8 \cdot \int (1 - sin^{2} (u)) cos (u) d u

Apply u-substitution

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

= 8 \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

= 8 (\int 1 d v - \int v^{2} d v)

\int 1 d v = v

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

= 8 (v - \frac{v ^{3}}{3})

Substitute back

= 8 (sin (\frac{x}{8}) - \frac{sin ^{3} ( \frac{x}{8} )}{3})

Add a constant to the solution

= 8 (sin (\frac{x}{8}) - \frac{sin ^{3} ( \frac{x}{8} )}{3}) + C

y = \frac{1}{x ^{5}}, (3, \frac{1}{243})

\int_{- 3}^{3} 18 - 2 x^{2} d x

x \to 2 - lim (\frac{3 - x}{2 - x})

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

x \to 0 + lim (\frac{1}{x})

What is the integral of cos^3(x/8) ?
The integral of cos^3(x/8) is 8(sin(x/8)-(sin^3(x/8))/3)+C