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

The integral of ((cos^3(x)))/3 is 1/3 (sin(x)-(sin^3(x))/3)+C

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

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

\frac{1}{3} (sin (x) - \frac{sin ^{3} ( x )}{3}) + C

Solution steps

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

Take the constant out:

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

= \frac{1}{3} \cdot \int cos^{3} (x) d x

Rewrite using trig identities

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

Apply u-substitution

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

= \frac{1}{3} \cdot \int 1 - u^{2} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{3} (\int 1 d u - \int u^{2} d u)

\int 1 d u = u

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

= \frac{1}{3} (u - \frac{u ^{3}}{3})

Substitute back

u = sin (x)

= \frac{1}{3} (sin (x) - \frac{sin ^{3} ( x )}{3})

Add a constant to the solution

= \frac{1}{3} (sin (x) - \frac{sin ^{3} ( x )}{3}) + C

x \to 3 lim ((2 x^{2} - 4 x - 8)^{6})

\frac{\partial}{\partial x} (\frac{4 x}{y})

\int \frac{4 x - 2}{x ^{3} - x ^{2} - 2 x} d x

y = sin (t^{3})

\int_{0}^{N} e^{- s t} e^{3 t} d t

What is the integral of ((cos^3(x)))/3 ?
The integral of ((cos^3(x)))/3 is 1/3 (sin(x)-(sin^3(x))/3)+C