hi
here,
∫cosx/tanx dx = ∫ cos2 x/ sinx dx
=∫1-sin2 x /sinx dx
=∫1/sinx dx – ∫ sinx dx
= ∫ cscx dx + cosx
∫ cscx dx = ∫ (csc2x +cotx cscx)/ (cscx+cotx)dx
u =cscx+cotx
du = -(cotxcscx+csc2 x) dx
= -∫1/u du
= -lnu = ln(1/u) = ln(1/(cscx+cotx))
∫cosx/tanx dx = ln(1/(cscx+cotx)) +cosx +C is the answer !