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

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

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integral of (sin(2x))/((1-cos(2x)))

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

\frac{1}{2} ln (2 sin^{2} (x)) + C

Solution steps

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

= \frac{1}{2} ln ∣ 1 - cos (2 x) ∣

Simplify

\frac{1}{2} ln ∣ 1 - cos (2 x) ∣ : \frac{1}{2} ln (2 sin^{2} (x))

= \frac{1}{2} ln (2 sin^{2} (x))

Add a constant to the solution

= \frac{1}{2} ln (2 sin^{2} (x)) + C

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