What is the integral of (4cos(2x))/(3-sin(2x)) ?

The integral of (4cos(2x))/(3-sin(2x)) is -2ln|3-sin(2x)|+C

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integral of (4cos(2x))/(3-sin(2x))

\int \frac{4 cos ( 2 x )}{3 - sin ( 2 x )} d x

- 2 ln ∣ 3 - sin (2 x) ∣ + C

Solution steps

\int \frac{4 cos ( 2 x )}{3 - sin ( 2 x )} d x

Take the constant out:

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

= 4 \cdot \int \frac{cos ( 2 x )}{3 - sin ( 2 x )} d x

Apply u-substitution

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

= 4 \cdot \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

= 4 (- \frac{1}{2} \cdot \int \frac{1}{u} d u)

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 4 (- \frac{1}{2} ln ∣ u ∣)

Substitute back

u = 3 - sin (2 x)

= 4 (- \frac{1}{2} ln ∣ 3 - sin (2 x) ∣)

Simplify

4 (- \frac{1}{2} ln ∣ 3 - sin (2 x) ∣) : - 2 ln ∣ 3 - sin (2 x) ∣

= - 2 ln ∣ 3 - sin (2 x) ∣

Add a constant to the solution

= - 2 ln ∣ 3 - sin (2 x) ∣ + C

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What is the integral of (4cos(2x))/(3-sin(2x)) ?
The integral of (4cos(2x))/(3-sin(2x)) is -2ln|3-sin(2x)|+C