What is the integral of tan^3(15pix) ?

The integral of tan^3(15pix) is 1/(15pi)(-ln|sec(15pix)|+(sec^2(15pix))/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(15pix)

\int tan^{3} (15 π x) d x

\frac{1}{15 π} (- ln ∣ sec (15 π x) ∣ + \frac{sec ^{2} ( 15 π x )}{2}) + C

Solution steps

\int tan^{3} (15 π x) d x

Apply u-substitution

: \int tan^{3} (u) \frac{1}{15 π} d u

= \int tan^{3} (u) \frac{1}{15 π} d u

Take the constant out:

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

= \frac{1}{15 π} \cdot \int tan^{3} (u) d u

Rewrite using trig identities

= \frac{1}{15 π} \cdot \int (- 1 + sec^{2} (u)) tan (u) d u

Apply u-substitution

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

= \frac{1}{15 π} \cdot \int \frac{- 1 + v ^{2}}{v} d v

Expand

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

= \frac{1}{15 π} \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}{15 π} (- \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}{15 π} (- ln ∣ v ∣ + \frac{v ^{2}}{2})

Substitute back

= \frac{1}{15 π} (- ln ∣ sec (15 π x) ∣ + \frac{sec ^{2} ( 15 π x )}{2})

Add a constant to the solution

= \frac{1}{15 π} (- ln ∣ sec (15 π x) ∣ + \frac{sec ^{2} ( 15 π x )}{2}) + C

\int_{1}^{4} \frac{4}{x} - \frac{16}{x ^{2}} d x

\int cos (\frac{x}{a}) sin (\frac{x}{a}) d x

\frac{d}{d x} (\frac{2 ^{x}}{1 - x ^{2}})

\int \frac{1}{x + !} d x

\frac{d}{d x} ((2 x - 1) (3 x + 4) (x + 7))

What is the integral of tan^3(15pix) ?
The integral of tan^3(15pix) is 1/(15pi)(-ln|sec(15pix)|+(sec^2(15pix))/2)+C