What is the integral of (tan^3(6/z))/(z^2) ?

The integral of (tan^3(6/z))/(z^2) is -1/6 (-ln|sec(6/z)|+(sec^2(6/z))/2)+C

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integral of (tan^3(6/z))/(z^2)

\int \frac{tan ^{3} ( \frac{6}{z} )}{z ^{2}} d z

- \frac{1}{6} (- ln sec (\frac{6}{z}) + \frac{sec ^{2} ( \frac{6}{z} )}{2}) + C

Solution steps

\int \frac{tan ^{3} ( \frac{6}{z} )}{z ^{2}} d z

Apply u-substitution

: \int - \frac{tan ^{3} ( u )}{6} d u

= \int - \frac{tan ^{3} ( u )}{6} d u

Take the constant out:

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

= - \frac{1}{6} \cdot \int tan^{3} (u) d u

Rewrite using trig identities

= - \frac{1}{6} \cdot \int (- 1 + sec^{2} (u)) tan (u) d u

Apply u-substitution

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

= - \frac{1}{6} \cdot \int \frac{- 1 + v ^{2}}{v} d v

Expand

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

= - \frac{1}{6} \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}{6} (- \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}{6} (- ln ∣ v ∣ + \frac{v ^{2}}{2})

Substitute back

= - \frac{1}{6} (- ln sec (\frac{6}{z}) + \frac{sec ^{2} ( \frac{6}{z} )}{2})

Add a constant to the solution

= - \frac{1}{6} (- ln sec (\frac{6}{z}) + \frac{sec ^{2} ( \frac{6}{z} )}{2}) + C

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What is the integral of (tan^3(6/z))/(z^2) ?
The integral of (tan^3(6/z))/(z^2) is -1/6 (-ln|sec(6/z)|+(sec^2(6/z))/2)+C