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

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

\int tan^{3} (3 x) d x

- \frac{1}{3} ln ∣ sec (3 x) ∣ + \frac{sec ^{2} ( 3 x )}{6} + C

Solution steps

\int tan^{3} (3 x) d x

Rewrite using trig identities

= \int (- 1 + sec^{2} (3 x)) tan (3 x) d x

Apply u-substitution

: \int \frac{- 1 + u ^{2}}{3 u} d u

= \int \frac{- 1 + u ^{2}}{3 u} d u

Expand

\frac{- 1 + u ^{2}}{3 u} : - \frac{1}{3 u} + \frac{u}{3}

= \int - \frac{1}{3 u} + \frac{u}{3} d u

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}{3 u} d u + \int \frac{u}{3} d u

\int \frac{1}{3 u} d u = \frac{1}{3} ln ∣ u ∣

\int \frac{u}{3} d u = \frac{u ^{2}}{6}

= - \frac{1}{3} ln ∣ u ∣ + \frac{u ^{2}}{6}

Substitute back

u = sec (3 x)

= - \frac{1}{3} ln ∣ sec (3 x) ∣ + \frac{sec ^{2} ( 3 x )}{6}

Add a constant to the solution

= - \frac{1}{3} ln ∣ sec (3 x) ∣ + \frac{sec ^{2} ( 3 x )}{6} + C

\frac{d}{d x} (ln (x) - e^{- x})

2 x y d y = (x^{2} - y^{2}) d x

\int e^{a x} cos (b x) d x

y = 4 - x^{2}, (- 2, 2)

\frac{\partial}{\partial x} (sin (4 x^{2} y - 4 x y^{3}))

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