What is the integral of tan(-6x+3) ?

The integral of tan(-6x+3) is 1/6 ln|cos(-6x+3)|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of tan(-6x+3)

\int tan (- 6 x + 3) d x

\frac{1}{6} ln ∣ cos (- 6 x + 3) ∣ + C

Solution steps

\int tan (- 6 x + 3) d x

Rewrite using trig identities

= \int \frac{sin ( - 6 x + 3 )}{cos ( - 6 x + 3 )} d x

Apply u-substitution

= \int \frac{1}{6 u} 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 \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{6} ln ∣ u ∣

Substitute back

u = cos (- 6 x + 3)

= \frac{1}{6} ln ∣ cos (- 6 x + 3) ∣

Add a constant to the solution

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

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