What is the integral of 5cos^3(6x) ?

The integral of 5cos^3(6x) is 5/6 (sin(6x)-1/3 sin^3(6x))+C

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integral of 5cos^3(6x)

\int 5 cos^{3} (6 x) d x

\frac{5}{6} (sin (6 x) - \frac{1}{3} sin^{3} (6 x)) + C

Solution steps

\int 5 cos^{3} (6 x) d x

Take the constant out:

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

= 5 \cdot \int cos^{3} (6 x) d x

Apply u-substitution

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

= 5 \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

= 5 \cdot \frac{1}{6} \cdot \int cos^{3} (u) d u

Rewrite using trig identities

= 5 \cdot \frac{1}{6} \cdot \int (1 - sin^{2} (u)) cos (u) d u

Apply u-substitution

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

= 5 \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

= 5 \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}

= 5 \cdot \frac{1}{6} (v - \frac{v ^{3}}{3})

Substitute back

= 5 \cdot \frac{1}{6} (sin (6 x) - \frac{sin ^{3} ( 6 x )}{3})

Simplify

5 \cdot \frac{1}{6} (sin (6 x) - \frac{sin ^{3} ( 6 x )}{3}) : \frac{5}{6} (sin (6 x) - \frac{1}{3} sin^{3} (6 x))

= \frac{5}{6} (sin (6 x) - \frac{1}{3} sin^{3} (6 x))

Add a constant to the solution

= \frac{5}{6} (sin (6 x) - \frac{1}{3} sin^{3} (6 x)) + C

\frac{\partial}{\partial x} (3 x^{3} + x^{2} y - 3 y^{3})

a x^{2}

\int 13 x^{2} sin (π x) d x

y = 10 x - x + 3, (100, 3)

\int_{0}^{10} 2 π y (100 - y^{2}) d y

What is the integral of 5cos^3(6x) ?
The integral of 5cos^3(6x) is 5/6 (sin(6x)-1/3 sin^3(6x))+C