Answer 1

`I = int_0^(pi/2) cos^2 x/(1 + sinxcosx)dx`   ....(1)

Using `int_0^a f(x) dx = int_0^a f(a -x) dx`

`I  = int_0^(pi/2) (cos^2 (pi/2 - x))/(1 + sin(pi/2 -x)cos(pi/2 -x)) dx`

`= int_0^(pi/2) (sin^2 x)/(1+cos x.sin x) dx`  .....(2)

Adding eq. (1) & (2)

`2I = int_0^(pi/2) (cos^2x + sin^2 x)/(1+sin xcos x) dx`

`= int_0^(pi/2) 1/(1+sinxcos x) dx``

`= int_0^(pi/2) (sec^2x) /(sec^2x + tan x) dx`

`2I = int_0^(pi/2) (sec^2 x dx)/(1+tan^2 x + tan x)`

Put `tan x = t, sec^2xdx = dt`

when x = 0, t = 0

when `s = pi/2, t = oo`

`2I = int_0^(oo) (dt)/(t^2 + 2t. 1/2+1/4 1/4 + 1)`

`= int_0^(oo)  (dt)/((t+1/2)^2 + ((sqrt3)/2)^2`

`= 1/(sqrt3/2) [tan^(-1) ((t+1/2)/(sqrt3/2))]_0^oo`

`= 2/sqrt3 tan^(-1) [(2t + 1)/sqrt3]_0^oo`

`2I = 2/sqrt3 [pi/2 - pi/6]`

`I = 1/sqrt3[(3pi - pi)/6]`

` = 1/sqrt3 [(2pi)/6] = pi/(3sqrt3)`