Question

How do you find the antiderivative of `cos(x)/(1-cos(x))`?

Answer 1

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

Explanation:

`I=intcos(x)/(1-cos(x))dx`

Rewriting the integral in a simpler form:

`I=int((cos(x)-1)+1)/(1-cos(x))dx`

`I=int(-(1-cos(x)))/(1-cos(x))dx+intdx/(1-cos(x))`

`I=-intdx+intdx/(1-cos(x))`

`I=-x+intdx/(1-cos(x))`

For the remaining integral, we'll use the tangent half-angle substitution which uses `t=tan(x/2)`. The function `cos(x)` can be expressed in terms of `tan(x/2)` as follows:

That is, `cos(x)=(1-tan^2(x/2))/(1+tan^2(x/2))=(1-tan^2(x/2))/sec^2(x/2)`.

Also note that the substitution `t=tan(x/2)` implies `dt=1/2sec^2(x/2)dx`.

Applying this to the integral gives:

`I=-x+intdx/((1-(1-tan^2(x/2))/sec^2(x/2)))`

`I=-x+int(sec^2(x/2)dx)/(sec^2(x/2)-(1-tan^2(x/2))`

Rewriting `sec^2(x/2)` as `tan^2(x/2)+1`:

`I=-x+int(sec^2(x/2)dx)/(2tan^2(x/2))`

`I=-x+2int(1/2sec^2(x/2)dx)/(2tan^2(x/2))`

Substituting:

`I=-x+2intdt/t^2`

`I=-x-2/t+C`

`I=-x-2/tan(x/2)+C`

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

Answer 2

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

Explanation:

`int cos x/(1-cos x) dx=int (1-2sin^2(x/2))/(2sin^2(x/2)) dx`

`=int1/2 csc^2(x/2) dx-int dx`

`=intd(-cot(x/2) -x`

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