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

The integral of 3cos^3(6x) is 1/2 (sin(6x)-1/3 sin^3(6x))+C

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integral of 3cos^3(6x)

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

\frac{1}{2} (sin (6 x) - \frac{1}{3} sin^{3} (6 x)) + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

= 3 \cdot \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

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

Rewrite using trig identities

= 3 \cdot \frac{1}{6} \cdot \int (1 - sin^{2} (u)) cos (u) d u

Apply u-substitution

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

= 3 \cdot \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

= 3 \cdot \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}

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

Substitute back

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

Simplify

3 \cdot \frac{1}{6} (sin (6 x) - \frac{sin ^{3} ( 6 x )}{3}) : \frac{1}{2} (sin (6 x) - \frac{1}{3} sin^{3} (6 x))

= \frac{1}{2} (sin (6 x) - \frac{1}{3} sin^{3} (6 x))

Add a constant to the solution

= \frac{1}{2} (sin (6 x) - \frac{1}{3} sin^{3} (6 x)) + C

\int cos^{10} (2 y) sin^{3} (2 y) d y

\frac{d}{d x} (\frac{1}{9} x - ln (3 x))

y = x^{2}, y = 8 x

y^{^{'}} - 6 x^{2} y = 0, y (1) = e

What is the integral of 3cos^3(6x) ?
The integral of 3cos^3(6x) is 1/2 (sin(6x)-1/3 sin^3(6x))+C