DR STRANGE
Last Activity: 8 Years ago
HI Alan,
here isthe solutuion,
∫ [ 1 / ( 1 + cos x/sin x ) ] dx
. = ∫ [ ( sin x ) / ( sin x + cos x ) ] dx
. = (1/2) • ∫ [ ( 2 sin x ) / ( sin x + cos x ) ] dx ...( Note This Step )
. = (1/2) • ∫ { [ ( sin x + cos x ) - ( cos x - sin x ) ] / ( sin x + cos x ) } dx ... This Too
. = (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 .................ans
plz approve if you are satisfied :)