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Evaluate : `(i) int_(0)^(pi//2)(cosx)/((cos (x)/(2)+sin (x)/(2)))dx` `(ii) int_(0)^(pi//2)(cosx)/((1+cosx+sinx))dx`

Answer» `(i) int_(0)^(pi//2)(cosx)/((cos.(x)/(2)+sin.(x)/(2)))dx=int_(0)^(pi//2)((cos^(2).(x)/(2)-sin^(2).(x)/(2)))/((cos.(x)/(2)+sin.(x)/(2)))dx`
`=int_(0)^(pi//2)(2(cos.(x)/(2)-sin.(x)/(2)))/((cos.(x)/(2)+sin.(x)/(2))^(2))dx=2*int_(1)^(sqrt(2))(dt)/(t^(2))=[(-2)/(t)]_(1)^(sqrt(2))=sqrt(2)(sqrt(2)-1)`.
[putting `cos.(x)/(2)+sin.(x)/(2)=t` and `(1)/(2)(cos.(x)/(2)-sin(x)/(2))dx=dt`,
also, `x=0impliest=1` and `x=(pi//2)impliest=sqrt(2)]`
`(ii)int_(0)^(pi//2)(cosx)/((1+cosx+sinx))dx = int_(0)^(pi//2)(cosx)/((1+cosx)+sinx)dx`
`=int_(0)^(pi//2)(cos^(2)(x//2)-sin^(2)(x//2))/([2cos^(2)(x//2)+2sin(x//2)cos(x//2)])dx`
`=(1)/(2)int_(0)^(pi//2)(1-tan^(2)(x//2))/(1+tan(x//2))dx` [dividing num. and denom. by `cos^(2)(x//2)`]
`=(1)/(2)int_(0)^(pi//2)[1-tan(x//2)]dx=(1)/(2)int_(0)^(pi//2)dx-(1)/(2)int_(0)^(pi//2)(sin(x//2))/(cos(x//2))dx`
`=(1)/(2)*[x]_(0)^(pi//2)+[logcos(x//2)]_(0)^(pi//2)`
`=(pi)/(4)+logcos.(pi)/(4)=(pi)/(4)+log((1)/(sqrt(2)))=((pi)/(4)-(1)/(2)log2)`.


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