What is the integral of 4cos^3(2x) ?

The integral of 4cos^3(2x) is 2(sin(2x)-1/3 sin^3(2x))+C

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integral of 4cos^3(2x)

\int 4 cos^{3} (2 x) d x

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

Solution steps

\int 4 cos^{3} (2 x) d x

Take the constant out:

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

= 4 \cdot \int cos^{3} (2 x) d x

Apply u-substitution

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

= 4 \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

= 4 \cdot \frac{1}{2} \cdot \int cos^{3} (u) d u

Rewrite using trig identities

= 4 \cdot \frac{1}{2} \cdot \int (1 - sin^{2} (u)) cos (u) d u

Apply u-substitution

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

= 4 \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

= 4 \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}

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

Substitute back

= 4 \cdot \frac{1}{2} (sin (2 x) - \frac{sin ^{3} ( 2 x )}{3})

Simplify

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

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

Add a constant to the solution

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

(x^{l n (x)})^{^{'}}

\frac{( s + 3 )}{( s ^{2} )}

\int cos (x) tan (x) d x

\int_{0}^{\frac{π}{3}} x tan^{2} (x) d x

\frac{d}{d x} (ln (\frac{1 + x}{1 - x}))

What is the integral of 4cos^3(2x) ?
The integral of 4cos^3(2x) is 2(sin(2x)-1/3 sin^3(2x))+C