Question

Find:

`int (x - sin x)/(1 - cos x) dx`

Answer 1

1 − cos x = 2 sin² `(x/2)`

sin x = 2 sin `(x/2) cos (x/2)`

`int (x - sin x)/(1 - cos x) dx`

`int (x - 2 sin (x/2) cos (x/2))/(2 sin^2 (x/2)) dx`

Now, split the fraction into two separate terms:

`int [x/(2 sin^2 (x/2)) - (2 sin (x/2) cos (x/2))/(2 sin^2 (x/2))] dx`

`1/2 int x csc^2 (x/2) dx - int cot (x/2)dx`

Solve the First Part using Integration by Parts

first integral: `int x csc^2 (x/2) dx`

u = x ⇒ du = dx

`dv = csc^2(x/2) dx = v = int csc^2 (x/2) dx = -2 cot (x/2)`

Using the formula ∫ udv = uv − ∫ v du:

`int x csc^2 (x/2) dx = x (-2 cot (x/2)) - int -2cot (x/2) dx`

= `-2x cot (x/2) + 2 int cot (x/2) dx`

Integration by Parts

= `1/2 [-2x cot (x/2) + 2 int cot (x/2) dx] - int cot (x/2)dx`

= `-x cot (x/2) + int cot (x/2) dx - int cot (x/2) dx + C`

= `-x cot (x/2) + C`