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

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

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integral of 5cos^3(3x)

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

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

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

Rewrite using trig identities

= 5 \cdot \frac{1}{3} \cdot \int (1 - sin^{2} (u)) cos (u) d u

Apply u-substitution

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

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

\int_{- \infty}^{1} x e^{x} d x

\int sin (x^{2} + 1) d x

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\frac{d y}{d x} = (sin^{2} (2 x)) (cos^{2} (y))

t \frac{d y}{d t} = t^{3} + 20 t^{3} y

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