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

The integral of 2tan^3(x/4) is 8(-ln|sec(x/4)|+1/2 sec^2(x/4))+C

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integral of 2tan^3(x/4)

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

8 (- ln sec (\frac{x}{4}) + \frac{1}{2} sec^{2} (\frac{x}{4})) + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

: \int tan^{3} (u) \cdot 4 d u

= 2 \cdot \int tan^{3} (u) \cdot 4 d u

Take the constant out:

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

= 2 \cdot 4 \cdot \int tan^{3} (u) d u

Rewrite using trig identities

= 2 \cdot 4 \cdot \int (- 1 + sec^{2} (u)) tan (u) d u

Apply u-substitution

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

= 2 \cdot 4 \cdot \int \frac{- 1 + v ^{2}}{v} d v

Expand

\frac{- 1 + v ^{2}}{v} : - \frac{1}{v} + v

= 2 \cdot 4 \cdot \int - \frac{1}{v} + v 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 4 (- \int \frac{1}{v} d v + \int v d v)

\int \frac{1}{v} d v = ln ∣ v ∣

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

= 2 \cdot 4 (- ln ∣ v ∣ + \frac{v ^{2}}{2})

Substitute back

= 2 \cdot 4 (- ln sec (\frac{x}{4}) + \frac{sec ^{2} ( \frac{x}{4} )}{2})

Simplify

2 \cdot 4 (- ln sec (\frac{x}{4}) + \frac{sec ^{2} ( \frac{x}{4} )}{2}) : 8 (- ln sec (\frac{x}{4}) + \frac{1}{2} sec^{2} (\frac{x}{4}))

= 8 (- ln sec (\frac{x}{4}) + \frac{1}{2} sec^{2} (\frac{x}{4}))

Add a constant to the solution

= 8 (- ln sec (\frac{x}{4}) + \frac{1}{2} sec^{2} (\frac{x}{4})) + C

n \to \infty lim ((1 + \frac{3}{n})^{- n^{2}})

\int 4 x e^{4 x} d x

x \to 0 lim (\frac{( tan ( x ) )}{x})

\frac{d}{d t} (e^{2 t s i n (2 t)})

\frac{d}{d x} (\frac{x ^{2}}{8} - \frac{1}{x})

What is the integral of 2tan^3(x/4) ?
The integral of 2tan^3(x/4) is 8(-ln|sec(x/4)|+1/2 sec^2(x/4))+C