If $\int\left[\ln\left(\frac{\cos 2\theta}{1+\sin 2\theta}\right) + \ln\left(\frac{1+\sin 2\theta}{1-\sin 2\theta}\right)^{\cos^2\theta}\right]d\theta$ is equal to $\frac{1}{a}\sin 2\theta\ln\left|\frac{\cos\theta + \sin\theta}{\cos\theta - \sin\theta}\right| + b\ln|\cos 2\theta| + c$ where $a, b \in \mathbb{R} - \{0\}$ & $c$ is integration constant such that $\cos\theta > \sin\theta > 0$ then $(a+b)$ is
Step-by-Step Solution
Key Concept: Logarithm properties decompose the fraction inside the logarithm, simplifying the IBP calculation.
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 $b = \frac{1}{2}$.
Correct Answer: Looking at the given solution, we have:
From the integration by parts result:
$$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