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

The integral of (sin(2x))/((1-cos(2x))^2) is -1/2 cot^2(2x)-1/8 cot^2(x)+1/8 tan^2(x)+C

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

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

- \frac{1}{2} cot^{2} (2 x) - \frac{1}{8} cot^{2} (x) + \frac{1}{8} tan^{2} (x) + C

Solution steps

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

Multiply by the conjugate of

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

= \int \frac{sin ( 2 x )}{( \frac{s i n ^{2} ( 2 x )}{1 + c o s ( 2 x )} ) ^{2}} d x

Expand

\frac{sin ( 2 x )}{( \frac{s i n ^{2} ( 2 x )}{1 + c o s ( 2 x )} ) ^{2}} : \frac{1}{sin ^{3} ( 2 x )} + \frac{2 cos ( 2 x )}{sin ^{3} ( 2 x )} + \frac{cos ^{2} ( 2 x )}{sin ^{3} ( 2 x )}

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int \frac{1}{sin ^{3} ( 2 x )} d x = \frac{1}{16} (- cot^{2} (x) + 4 ln ∣ tan (x) ∣ + tan^{2} (x))

\int \frac{2 cos ( 2 x )}{sin ^{3} ( 2 x )} d x = - \frac{1}{2} cot^{2} (2 x)

\int \frac{cos ^{2} ( 2 x )}{sin ^{3} ( 2 x )} d x = - \frac{1}{4} ln ∣ tan (x) ∣ - \frac{1}{16} cot^{2} (x) + \frac{1}{16} tan^{2} (x)

= \frac{1}{16} (- cot^{2} (x) + 4 ln ∣ tan (x) ∣ + tan^{2} (x)) - \frac{1}{2} cot^{2} (2 x) - \frac{1}{4} ln ∣ tan (x) ∣ - \frac{1}{16} cot^{2} (x) + \frac{1}{16} tan^{2} (x)

Simplify

\frac{1}{16} (- cot^{2} (x) + 4 ln ∣ tan (x) ∣ + tan^{2} (x)) - \frac{1}{2} cot^{2} (2 x) - \frac{1}{4} ln ∣ tan (x) ∣ - \frac{1}{16} cot^{2} (x) + \frac{1}{16} tan^{2} (x) : - \frac{1}{2} cot^{2} (2 x) - \frac{1}{8} cot^{2} (x) + \frac{1}{8} tan^{2} (x)

= - \frac{1}{2} cot^{2} (2 x) - \frac{1}{8} cot^{2} (x) + \frac{1}{8} tan^{2} (x)

Add a constant to the solution

= - \frac{1}{2} cot^{2} (2 x) - \frac{1}{8} cot^{2} (x) + \frac{1}{8} tan^{2} (x) + C

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What is the integral of (sin(2x))/((1-cos(2x))^2) ?
The integral of (sin(2x))/((1-cos(2x))^2) is -1/2 cot^2(2x)-1/8 cot^2(x)+1/8 tan^2(x)+C