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

The integral of cos^5(3y) is 1/3 (sin(3y)-(2sin^3(3y))/3+(sin^5(3y))/5)+C

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integral of cos^5(3y)

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

\frac{1}{3} (sin (3 y) - \frac{2 sin ^{3} ( 3 y )}{3} + \frac{sin ^{5} ( 3 y )}{5}) + C

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

Rewrite using trig identities

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

Apply u-substitution

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

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

Expand

(1 - v^{2})^{2} : 1 - 2 v^{2} + v^{4}

= \frac{1}{3} \cdot \int 1 - 2 v^{2} + v^{4} 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}{3} (\int 1 d v - \int 2 v^{2} d v + \int v^{4} d v)

\int 1 d v = v

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

\int v^{4} d v = \frac{v ^{5}}{5}

= \frac{1}{3} (v - \frac{2 v ^{3}}{3} + \frac{v ^{5}}{5})

Substitute back

= \frac{1}{3} (sin (3 y) - \frac{2 sin ^{3} ( 3 y )}{3} + \frac{sin ^{5} ( 3 y )}{5})

Add a constant to the solution

= \frac{1}{3} (sin (3 y) - \frac{2 sin ^{3} ( 3 y )}{3} + \frac{sin ^{5} ( 3 y )}{5}) + C

x \to 0 lim (\frac{cos ( x )}{x})

y \to 0 lim (sin (\frac{1}{y}))

\frac{d}{d x} (tan (e^{x}) \cdot cosh (e^{x}))

\frac{d ^{4}}{d x ^{4}} (ln (x + 1))

\frac{d y}{d x} = (x + 1)

What is the integral of cos^5(3y) ?
The integral of cos^5(3y) is 1/3 (sin(3y)-(2sin^3(3y))/3+(sin^5(3y))/5)+C