Integrate (1)/1+cotx using substitution method.

∫ [ 1 / ( 1 + cos x/sin x ) ] dx

= ∫ [ ( sin x ) / ( sin x + cos x ) ] dx

= (1/2) • ∫ [ ( 2 sin x ) / ( sin x + cos x ) ] dx …

= (1/2) • ∫ { [ ( sin x + cos x ) – ( cos x – sin x ) ] / ( sin x + cos x ) } dx …

= (1/2) • { ∫ 1 dx – ∫ [ ( cos x – sin x ) / ( sin x + cos x ) ] dx }

= (1/2) • { x – ∫ (1/u) du }, …

where … u = sin x + cos x

= ( x/2 ) – (1/2)· ln | u | + C

= ( x/2 ) – (1/2)· ln | sin x + cos x | + C