What is the integral of (tan^3(ln(x)))/(9x) ?

The integral of (tan^3(ln(x)))/(9x) is 1/9 (-ln|sec(ln(x))|+(sec^2(ln(x)))/2)+C

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integral of (tan^3(ln(x)))/(9x)

\int \frac{tan ^{3} ( ln ( x ) )}{9 x} d x

\frac{1}{9} (- ln ∣ sec (ln (x)) ∣ + \frac{sec ^{2} ( ln ( x ) )}{2}) + C

Solution steps

\int \frac{tan ^{3} ( ln ( x ) )}{9 x} d x

Take the constant out:

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

= \frac{1}{9} \cdot \int \frac{tan ^{3} ( ln ( x ) )}{x} d x

Apply u-substitution

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

= \frac{1}{9} \cdot \int tan^{3} (u) d u

Rewrite using trig identities

= \frac{1}{9} \cdot \int (- 1 + sec^{2} (u)) tan (u) d u

Apply u-substitution

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

= \frac{1}{9} \cdot \int \frac{- 1 + v ^{2}}{v} d v

Expand

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

= \frac{1}{9} \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

= \frac{1}{9} (- \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}

= \frac{1}{9} (- ln ∣ v ∣ + \frac{v ^{2}}{2})

Substitute back

= \frac{1}{9} (- ln ∣ sec (ln (x)) ∣ + \frac{sec ^{2} ( ln ( x ) )}{2})

Add a constant to the solution

= \frac{1}{9} (- ln ∣ sec (ln (x)) ∣ + \frac{sec ^{2} ( ln ( x ) )}{2}) + C

\frac{d}{d x} (5 - 6 x)

\int_{1}^{5 e} \frac{1}{x} d x

n = 1 \sum \infty \frac{n}{5 n + 1}

ln (\frac{k}{l})

\int e^{t} e^{t} d t

What is the integral of (tan^3(ln(x)))/(9x) ?
The integral of (tan^3(ln(x)))/(9x) is 1/9 (-ln|sec(ln(x))|+(sec^2(ln(x)))/2)+C