Pls find the integral of cot 2x

= ∫ cot(2x) dx 

Let: 
u = 2x 
du = 2 dx 
(1/2)du = dx 

Remember this identity: cotθ = cosθ/sinθ 
= ∫ cot(u) * (1/2)du 
= 1/2 ∫ cot(u) du 
= 1/2 ∫ cos(u)/sin(u) du 

Let: 
w = sin(u) 
dw = cos(u) du 

= 1/2 ∫ dw / w 
= 1/2 ∫ 1/w dw 
= 1/2ln|w| + C (Back-substitute) 
= 1/2ln|sin(u)| + C (Back-substitute again) 
= 1/2ln|sin(2x)| + C (Answer) 

Hope this helps!