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Discussion Response Post to: definite

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puneet (2841)

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

= I₁ + I₂ .. (say)

Now I₁ = ₀^/2cosx/(2+cosx) dx

= [sinx/(2+cosx)]₀^/2- ₀^/2sinx.[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

Puneet Agrawal IIT Delhi

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