hiiiii liku

u have asked some really tough ones ...... after a long time i had to use do much brain .... :) never mind, take a salute from me and let us see how to solve it ..

 

Now we have I =  ₀^/2 1+2cosx/(2+cosx)² dx 

 

now i will use a simple manipulation .. how i reached it might be a question that u have ... well just observe the 2 in denominator .. that 2 suggests this manipulation shud be made ...

 

I =  ₀^/2  1+2cosx/(2+cosx)² dx

 = ₀^/2 [(cos²x + sin²x) + 2cosx]/(2+cosx)² dx

 =  ₀^/2 [(cosx(cosx +  2) + sin²x)]/(2+cosx)² dx

 = ₀^/2 cosx/(2+cosx) dx +  ₀^/2 sin²x/(2+cosx)² dx

 

= I₁ + I₂           ..  (say)

 

Now I₁  = ₀^/2 cosx/(2+cosx) dx  

           = [sinx/(2+cosx)]₀^/2  -  ₀^/2 sinx.[sinx/(2+cosx)²] dx (using integration by parts)

 

          = [sinx/(2+cosx)]₀^/2  - I₂

 

so, we get I₁ + I₂  =  [sinx/(2+cosx)]₀^/2 

or, I = [sinx/(2+cosx)]₀^/2 

         = 1/2 - 0

       = 1/2

 

Thus ₀^/2 1+2cosx/(2+cosx)² dx = 1/2 ans

 

cheers