Question Number 102303 by bemath last updated on 08/Jul/20

∫ ((1+csc 2x)/(1−sin 2x)) dx ?

Commented bybemath last updated on 08/Jul/20

thank you both

Answered by 1549442205 last updated on 08/Jul/20

F=∫(1/(1−sin2x))dx+∫((cos2x)/((cosx−sinx)^2 ))dx ∫(dx/((cosx−sinx)^2 ))+∫((cosx+sinx)/(cosx−sinx))dx =∫(dx/((1/2)cos^2 (x+(π/4))))+∫((((√2)/2)sin(x+(π/4))dx)/(((√2)/2)cos(x+(π/4)))) =2tan(x+(π/4))+∫tan(x+(π/4))dx =2tan(x+(𝛑/4))+ln∣cos(x+(𝛑/4))∣+C

Answered by Dwaipayan Shikari last updated on 08/Jul/20

∫(1/(1−sin2x))−(1/2)∫((−2cos2x)/(1−sin2x))dx ∫(1/(1−((2t)/(t^2 +1)))).(1/(t^2 +1))dt−(1/2)log(1−sin2x) {put tanx=t ∫(1/((t−1)^2 ))dt−(1/2)log(1−sin2x)=(1/(1−t))−(1/2)log(1−sin2x) =((cosx)/(cosx−sinx))−(1/2)log(1−sin2x)+C