Question

Integration of secx/(1+cosecx)

Answer 1

this is the solution please Solve the next step because pics of answer is not send by me sorry

Answer 2

 I = ∫ [ 1 / ( sec x + cos x ) ] dx 

..= ∫ { 1 / [ (1/cos x) + (1/sin x) ] } dx 

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

..= 2 ∫ [ ( cos x + sin x ) / ( 2 sin x. cos x ) ] dx 

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

..= 2 ∫ { ( cos x + sin x ) / [ 1 - ( sin² x + cos² x - 2 sin x. cos x ) ] } dx 

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

..= 2 ∫ [ 1 ( 1 - u² ) ] du, ......... where ... u = sin x - cos x 

..= ∫ { [ 1 / (1+u) ] + [ 1 / (1-u) ] } du ..... Partial Fractions 

..= ln | 1+u | - ln | 1-u | + C 

..= ln | (1+u) / (1-u) | + C 

..= ln | ( 1 + sin x - cos x ) / ( 1 - sin x + cos x ) | + C .