The correct options are
A √2log|cot3π8−x4|+c
B √2log|tan(π8+x4)|+c
C √2log|sec(π4−x2)+tan(π4−x2)|+c
D √2log|cosec(π4+x2)−cot(π4+x2)|+c
Given ∫dx(1+sinx)1/2
⇒ ∫dx(cos(x/2)+sin(x/2))2×(1/2)
⇒ ∫dx((1/√2)cos(x/2)+(1/√2)sin(x/2))(√2)
⇒ ∫dx(sin(π/4)cos(x/2)+(cos(π/4)sin(x/2))(√2)
⇒ ∫dx(sin((π/4)+(x/2))(√2)
⇒ ∫1√2cosec((x/2)+(π/4))dx
Since ∫cosec(x)dx=log|cosec(x)−cot(x)|
Hence (√2)log|cosec((x/2)+(π/4))−cot((x/2)+(π/4))|+c ie option B is the answer.