What is the integral of (sin(3x))/(1+cos(3x)) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of (sin(3x))/(1+cos(3x))

\int \frac{sin ( 3 x )}{1 + cos ( 3 x )} d x

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

Solution steps

\int \frac{sin ( 3 x )}{1 + cos ( 3 x )} d x

Apply u-substitution

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - \frac{1}{3} ln ∣ u ∣

Substitute back

u = 1 + cos (3 x)

= - \frac{1}{3} ln ∣ 1 + cos (3 x) ∣

Add a constant to the solution

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

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