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

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

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integral of 2cos^3(2x)

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

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

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

: \int cos^{3} (u) \frac{1}{2} d u

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

Take the constant out:

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

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

Rewrite using trig identities

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

Apply u-substitution

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

= 2 \cdot \frac{1}{2} \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

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

\int 1 d v = v

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

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

sin (θ) + csc (θ)

\frac{d}{d x} (\frac{a ^{2} + x ^{2}}{a ^{2} - x ^{2}})

\int_{- 1}^{0} 2 - e^{- x} d x

\frac{x}{x ^{2} + 4}

\frac{d}{d x} (\frac{x - 3}{x + 3})

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