Question

How do you evaluate the integral `int dx/(1-cos3x)`?

Answer 1

`int dx/(1-cos3x) = -1/3cot((3x)/2)+C`

Explanation:

Substitute `t=3x`:

`int dx/(1-cos3x) = 1/3 int (dt)/(1-cost)`

Use now the trigonometric identity:

`cost= (1-tan^2(t/2))/(1+tan^2(t/2))`

So that:

`1/(1-cost) = 1/(1-(1-tan^2(t/2))/(1+tan^2(t/2)))`

`1/(1-cost) = (1+tan^2(t/2))/((1+tan^2(t/2))-(1-tan^2(t/2)))`

`1/(1-cost) = (1+tan^2(t/2))/(1+tan^2(t/2)-1+tan^2(t/2))`

`1/(1-cost) = (1+tan^2(t/2))/(2tan^2(t/2))`

And as:

`1+tan^2alpha = 1+ sin^2alpha/cos^2alpha = (cos^2alpha+sin^2alpha)/cos^2alpha = 1/cos^2alpha = sec^2alpha`

we get:

`int (dt)/(1-cost) = int sec^2(t/2)/(2tan^2(t/2))dt`

Substitute now:

`u=tan(t/2)`

`du=1/2sec^2(t/2)dt`

and we have:

`int sec^2(t/2)/(2tan^2(t/2))dt = int (du)/u^2 = -1/u+C`

and undoing the substitution:

`int (dt)/(1-cost) = -1/tan(t/2)+C = -cot(t/2)+C`

`int dx/(1-cos3x) = -1/3cot((3x)/2)+C`