\int_{-\pi /2}^{\pi /2}{log\left[\frac{ax^{2}+bx+c}{ax^{2}-bx+c} (a+b)\left|sinx \right|\right]}dx
ans---------->\pi ln\left(\frac{a+b}{2} \right)
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1 Answers
29log\frac{ax^{2} + bx +c}{ax^{2} - bx +c} (a +b ) |sinx| = log \frac{ax^{2} + bx +c}{ax^{2} - bx +c} +log (a+b) + log|sinx|
\int_{-\pi /2}^{\pi /2} log \frac{ax^{2} + bx +c}{ax^{2} - bx +c} +\int_{-\pi /2}^{\pi /2}log (a+b) + \int_{-\pi /2}^{\pi /2}log|sinx| =
0 + \pi log(a+b) + \pi log\frac{1}{2} = \pi log\frac{(a+b)}{2}
PS : log\frac{ax^{2} + bx +c}{ax^{2} - bx +c} is an odd function
