What is the integral of (sin(2x))/(1+3cos(2x)) ?

The integral of (sin(2x))/(1+3cos(2x)) is -1/6 ln|1+3cos(2x)|+C

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integral of (sin(2x))/(1+3cos(2x))

\int \frac{sin ( 2 x )}{1 + 3 cos ( 2 x )} d x

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

Solution steps

\int \frac{sin ( 2 x )}{1 + 3 cos ( 2 x )} 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 = 1 + 3 cos (2 x)

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

Add a constant to the solution

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

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