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

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

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integral of cos^5(3x)

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

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

Solution steps

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

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 x) - \frac{2 sin ^{3} ( 3 x )}{3} + \frac{sin ^{5} ( 3 x )}{5})

Add a constant to the solution

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

x \to 0 + lim (\frac{x + 97}{x})

\frac{d}{d w} (lw)

\int x^{5} + \frac{1}{x ^{7}} d x

sin (100 t + 30)

\frac{d}{d x} f (x) = 2 x - 7

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