Evaluate: `int_(-pi/2)^(pi/2)log{(a x^2+b x+c)/(a x^2-b x+c)(a+b)|sinx|}dx`

Explanation

`I=int_(-pi//2)^(pi//2) log {(ax^(2)+bx+c)/(ax^(2)-bx+c)(a+b)|sinx|}dx`
`=int_(-pi//2)^(pi//2) log((ax^(2)+bx+c)/(ax^(2)-bx+c))dx+int_(-pi//2)^(pi//2) log (a+b)dx`
`+int_(-pi//2)^(pi//2) log|sinx|dx`……………1
`=I_(1)+I_(2)+I_(3)`
Now let `f(x)=log((ax^(2)+bx+c)/(ax^(2)-bx+c))`
or `f(-x)=log((ax^(2)-bx+c)/(ax^(2)+bx+c))`
`=-f(x)`
`:.I_(1)=int_(-pi//2)^(pi//2) f(x)dx=0`
`I_(2)=log(a+b)[x]_(-pi//2)^(pi//2)`
`=pi log(a+b)`
`I_(3)=int_(-pi//2)^(pi//2) log|sinx|dx`
`=2int_(0)^(pi//2)log|sinx|dx`
`=2int_(0)^(pi//2)logsinx dx`
`=2(-1/2pi log 2)`
Hence, from 1 we have
`I=0+pi log(a+b)-pi log 2`
`=pi log{(a+b)//2}`