What is the integral of (tan(ln(3x+4)))/(3x+4) ?

The integral of (tan(ln(3x+4)))/(3x+4) is -1/3 ln|cos(ln(3x+4))|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (tan(ln(3x+4)))/(3x+4)

\int \frac{tan ( ln ( 3 x + 4 ) )}{3 x + 4} d x

- \frac{1}{3} ln ∣ cos (ln (3 x + 4)) ∣ + C

Solution steps

\int \frac{tan ( ln ( 3 x + 4 ) )}{3 x + 4} d x

Apply u-substitution

: \int \frac{tan ( ln ( u ) )}{3 u} d u

= \int \frac{tan ( ln ( u ) )}{3 u} d u

Take the constant out:

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

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

Apply u-substitution

: \int tan (v) d v

= \frac{1}{3} \cdot \int tan (v) d v

Rewrite using trig identities

= \frac{1}{3} \cdot \int \frac{sin ( v )}{cos ( v )} d v

Apply u-substitution

: \int - \frac{1}{w} d w

= \frac{1}{3} \cdot \int - \frac{1}{w} d w

Take the constant out:

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

= \frac{1}{3} (- \int \frac{1}{w} d w)

Use the common integral:

\int \frac{1}{w} d w = ln (∣ w ∣)

= \frac{1}{3} (- ln ∣ w ∣)

Substitute back

= \frac{1}{3} (- ln ∣ cos (ln (3 x + 4)) ∣)

Simplify

= - \frac{1}{3} ln ∣ cos (ln (3 x + 4)) ∣

Add a constant to the solution

= - \frac{1}{3} ln ∣ cos (ln (3 x + 4)) ∣ + C

\int \frac{1}{cos ^{6} ( x )} d x

x \to \infty lim (\frac{( 5 + 2 e ^{- 2 x} )}{ln ( x )})

17 - x^{2}

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

4 - x^{2}, x = 0

What is the integral of (tan(ln(3x+4)))/(3x+4) ?
The integral of (tan(ln(3x+4)))/(3x+4) is -1/3 ln|cos(ln(3x+4))|+C