What is the integral of (tan^3(9/z))/(z^2) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (tan^3(9/z))/(z^2)

\int \frac{tan ^{3} ( \frac{9}{z} )}{z ^{2}}

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

Solution steps

\int \frac{tan ^{3} ( \frac{9}{z} )}{z ^{2}} d z

Apply u-substitution

: \int - \frac{tan ^{3} ( u )}{9} d u

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

Take the constant out:

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

= - \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 (\frac{9}{z}) + \frac{sec ^{2} ( \frac{9}{z} )}{2})

Add a constant to the solution

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

\frac{d}{d x} ((lo g_{2} (x))^{2})

x \to 1 - lim (\frac{- 3 x - 1}{x - 1})

x \to \frac{π}{2} lim (x - \frac{π}{2})

\frac{\partial}{\partial z} (x^{2} + yz)

f (x) = x^{2} + 2 x, at x = 1

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