What is the integral of tan^5(x/2) ?

The integral of tan^5(x/2) is 2(ln|sec(x/2)|-sec^2(x/2)+(sec^4(x/2))/4)+C

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integral of tan^5(x/2)

\int tan^{5} (\frac{x}{2}) d x

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

Solution steps

\int tan^{5} (\frac{x}{2}) d x

Apply u-substitution

: \int tan^{5} (u) \cdot 2 d u

= \int tan^{5} (u) \cdot 2 d u

Take the constant out:

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

= 2 \cdot \int tan^{5} (u) d u

Rewrite using trig identities

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

Apply u-substitution

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

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

Expand

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

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

\int \frac{1}{v} d v = ln ∣ v ∣

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

\int v^{3} d v = \frac{v ^{4}}{4}

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

Substitute back

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

Add a constant to the solution

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

\frac{d}{d x} (\frac{( x ^{7} )}{x ^{14} + 1})

\int ln (x^{2} + a^{2}) d x

\frac{\partial}{\partial x} (x + i y)

\int \frac{( x - 1 )}{x ^{3} - x ^{2} - 2 x} d x

2 - x

What is the integral of tan^5(x/2) ?
The integral of tan^5(x/2) is 2(ln|sec(x/2)|-sec^2(x/2)+(sec^4(x/2))/4)+C