Advertisement Remove all ads

Evaluate `Int_0^(Pi/2) Cos^2x/(1+ Sinx Cosx) Dx` - Mathematics

Evaluate `int_0^(pi/2) cos^2x/(1+ sinx cosx) dx`

Advertisement Remove all ads

Solution

`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)`

  Is there an error in this question or solution?
Advertisement Remove all ads

APPEARS IN

Advertisement Remove all ads
Advertisement Remove all ads
Share
Notifications

View all notifications


      Forgot password?
View in app×