What is the integral of cos^5(2t) ?

The integral of cos^5(2t) is 1/2 sin(2t)-1/3 sin^3(2t)+1/10 sin^5(2t)+C

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integral of cos^5(2t)

\int cos^{5} (2 t) d t

\frac{1}{2} sin (2 t) - \frac{1}{3} sin^{3} (2 t) + \frac{1}{10} sin^{5} (2 t) + C

Solution steps

\int cos^{5} (2 t) d t

Apply u-substitution

: \int cos^{5} (u) \frac{1}{2} d u

= \int cos^{5} (u) \frac{1}{2} d u

Take the constant out:

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

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

Rewrite using trig identities

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

Apply u-substitution

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

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

Expand

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

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

Substitute back

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

Simplify

\frac{1}{2} (sin (2 t) - \frac{2 sin ^{3} ( 2 t )}{3} + \frac{sin ^{5} ( 2 t )}{5}) : \frac{1}{2} sin (2 t) - \frac{1}{3} sin^{3} (2 t) + \frac{1}{10} sin^{5} (2 t)

= \frac{1}{2} sin (2 t) - \frac{1}{3} sin^{3} (2 t) + \frac{1}{10} sin^{5} (2 t)

Add a constant to the solution

= \frac{1}{2} sin (2 t) - \frac{1}{3} sin^{3} (2 t) + \frac{1}{10} sin^{5} (2 t) + C

\frac{d}{d x} \frac{x}{1 - ( ln ( x ) ) ^{2}}

\int cos (x^{2} - 5) d x

x \to 2 lim (\frac{3 ^{x}}{x + 1})

\int x e^{x + 1} d x

\int sec^{3} (v) d v

What is the integral of cos^5(2t) ?
The integral of cos^5(2t) is 1/2 sin(2t)-1/3 sin^3(2t)+1/10 sin^5(2t)+C