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

The integral of cos^3(6x) is 1/6 (sin(6x)-(sin^3(6x))/3)+C

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integral of cos^3(6x)

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

\frac{1}{6} (sin (6 x) - \frac{sin ^{3} ( 6 x )}{3}) + C

Solution steps

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

Apply u-substitution

: \int cos^{3} (u) \frac{1}{6} d u

= \int cos^{3} (u) \frac{1}{6} d u

Take the constant out:

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

= \frac{1}{6} \cdot \int cos^{3} (u) d u

Rewrite using trig identities

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

Apply u-substitution

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

= \frac{1}{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

= \frac{1}{6} (\int 1 d v - \int v^{2} d v)

\int 1 d v = v

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

= \frac{1}{6} (v - \frac{v ^{3}}{3})

Substitute back

= \frac{1}{6} (sin (6 x) - \frac{sin ^{3} ( 6 x )}{3})

Add a constant to the solution

= \frac{1}{6} (sin (6 x) - \frac{sin ^{3} ( 6 x )}{3}) + C

x + y = 0, x + y^{2} = 56

x^{n}

\frac{d}{d x} (x^{7} x)

f (x) = \frac{( x ^{2} + 8 )}{7 x - 1}

\int \frac{x}{6 x ^{2} + 1} d x

What is the integral of cos^3(6x) ?
The integral of cos^3(6x) is 1/6 (sin(6x)-(sin^3(6x))/3)+C