What is the integral of cos^3(x/(17)) ?

The integral of cos^3(x/(17)) is 17(sin(x/(17))-(sin^3(x/(17)))/3)+C

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integral of cos^3(x/(17))

\int cos^{3} (\frac{x}{17}) d x

17 (sin (\frac{x}{17}) - \frac{sin ^{3} ( \frac{x}{17} )}{3}) + C

Solution steps

\int cos^{3} (\frac{x}{17}) d x

Apply u-substitution

: \int cos^{3} (u) \cdot 17 d u

= \int cos^{3} (u) \cdot 17 d u

Take the constant out:

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

= 17 \cdot \int cos^{3} (u) d u

Rewrite using trig identities

= 17 \cdot \int (1 - sin^{2} (u)) cos (u) d u

Apply u-substitution

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

= 17 \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

= 17 (\int 1 d v - \int v^{2} d v)

\int 1 d v = v

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

= 17 (v - \frac{v ^{3}}{3})

Substitute back

= 17 (sin (\frac{x}{17}) - \frac{sin ^{3} ( \frac{x}{17} )}{3})

Add a constant to the solution

= 17 (sin (\frac{x}{17}) - \frac{sin ^{3} ( \frac{x}{17} )}{3}) + C

\frac{d y}{d x}, sin (x y^{3}) = y^{3}

\int \frac{1 - x ^{2}}{4} d x

cos^{3} (x), at x = 0

\frac{2 x ^{3}}{3} - 12 x

\frac{d}{d x} (\frac{x ^{2}}{x + 5})

What is the integral of cos^3(x/(17)) ?
The integral of cos^3(x/(17)) is 17(sin(x/(17))-(sin^3(x/(17)))/3)+C