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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

Rewrite using trig identities

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

Apply u-substitution

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

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

Expand

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

= \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

= \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}

= ln ∣ v ∣ - v^{2} + \frac{v ^{4}}{4}

Substitute back

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

Add a constant to the solution

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

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