What is the integral of tan^5(x/(14)) ?

The integral of tan^5(x/(14)) is 14(ln|sec(x/(14))|-sec^2(x/(14))+(sec^4(x/(14)))/4)+C

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integral of tan^5(x/(14))

\int tan^{5} (\frac{x}{14}) d x

14 (ln sec (\frac{x}{14}) - sec^{2} (\frac{x}{14}) + \frac{sec ^{4} ( \frac{x}{14} )}{4}) + C

Solution steps

\int tan^{5} (\frac{x}{14}) d x

Apply u-substitution

: \int tan^{5} (u) \cdot 14 d u

= \int tan^{5} (u) \cdot 14 d u

Take the constant out:

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

= 14 \cdot \int tan^{5} (u) d u

Rewrite using trig identities

= 14 \cdot \int (- 1 + sec^{2} (u))^{2} tan (u) d u

Apply u-substitution

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

= 14 \cdot \int \frac{( - 1 + v ^{2} ) ^{2}}{v} d v

Expand

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

= 14 \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

= 14 (\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}

= 14 (ln ∣ v ∣ - v^{2} + \frac{v ^{4}}{4})

Substitute back

= 14 (ln sec (\frac{x}{14}) - sec^{2} (\frac{x}{14}) + \frac{sec ^{4} ( \frac{x}{14} )}{4})

Add a constant to the solution

= 14 (ln sec (\frac{x}{14}) - sec^{2} (\frac{x}{14}) + \frac{sec ^{4} ( \frac{x}{14} )}{4}) + C

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What is the integral of tan^5(x/(14)) ?
The integral of tan^5(x/(14)) is 14(ln|sec(x/(14))|-sec^2(x/(14))+(sec^4(x/(14)))/4)+C