Question:

We have $I = \int 2\cos^2\theta \ln\left(\frac{\cos\theta + \sin\theta}{\cos\theta - \sin\theta}\right) d\theta$. Using integration by parts with $u = \ln\left(\frac{\cos\theta + \sin\theta}{\cos\theta - \sin\theta}\right)$ and $dv = 2\cos^2\theta d\theta$, we simplify using $\ln\left(\frac{\cos\theta + \sin\theta}{\cos\theta - \sin\theta}\right) = \ln(\cos\theta + \sin\theta) - \ln(\cos\theta - \sin\theta)$. After applying IBP, the result is $I = \frac{\sin 2\theta}{2}\ln\left(\frac{\cos\theta + \sin\theta}{\cos\theta - \sin\theta}\right) + \frac{1}{2}\ln|\cos 2\theta| + c$, giving $a = 2$ and $b = \frac{1}{2}$.