What is the integral of 3cos^5(2x) ?

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

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integral of 3cos^5(2x)

\int 3 cos^{5} (2 x) d x

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

Solution steps

\int 3 cos^{5} (2 x) d x

Take the constant out:

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

= 3 \cdot \int cos^{5} (2 x) d x

Apply u-substitution

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

= 3 \cdot \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

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

Rewrite using trig identities

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

Apply u-substitution

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

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

Expand

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

= 3 \cdot \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

= 3 \cdot \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}

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

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