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

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

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

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

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

Solution steps

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

Apply u-substitution

: \int - \frac{1}{2 u} d u

= \int - \frac{1}{2 u} d u

Take the constant out:

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

= - \frac{1}{2} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - \frac{1}{2} ln ∣ u ∣

Substitute back

u = 3 + cos (2 x)

= - \frac{1}{2} ln ∣ 3 + cos (2 x) ∣

Add a constant to the solution

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

1 + 6 t

\frac{d}{d x} (\frac{x ^{2} + 16}{x})

\frac{d}{d x} (\frac{x ^{3} - x - 2}{- 2 x})

f (x) = \frac{40 x ^{5}}{e ^{x}}

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