Key Concepts and Formulas

• Trigonometric transformations: $a \cos x + b \sin x = R \cos(x - \alpha)$, where $R = \sqrt{a^2 + b^2}$ and $\tan \alpha = \frac{-b}{a}$.
• Integral of secant function: $\int \sec x \, dx = \ln |\sec x + \tan x| + C = \ln \left| \tan \left( \frac{\pi}{4} + \frac{x}{2} \right) \right| + C$.

Step-by-Step Solution

Step 1: Rewrite the integrand using trigonometric identities. We want to express $\cos x - \sin x$ in the form $R\cos(x + \alpha)$. Here, we have $1 \cdot \cos x + (-1) \cdot \sin x$. Therefore, $R = \sqrt{1^2 + (-1)^2} = \sqrt{2}$. We also have $\cos x - \sin x = \sqrt{2} \left( \frac{1}{\sqrt{2}} \cos x - \frac{1}{\sqrt{2}} \sin x \right) = \sqrt{2} \left( \cos \frac{\pi}{4} \cos x - \sin \frac{\pi}{4} \sin x \right) = \sqrt{2} \cos \left( x + \frac{\pi}{4} \right)$. Thus, the integral becomes $\int \frac{dx}{\cos x - \sin x} = \int \frac{dx}{\sqrt{2} \cos \left(x + \frac{\pi}{4} \right)}$

Step 2: Simplify the integral. We can rewrite the integral as: $\frac{1}{\sqrt{2}} \int \frac{dx}{\cos \left(x + \frac{\pi}{4} \right)} = \frac{1}{\sqrt{2}} \int \sec \left(x + \frac{\pi}{4} \right) dx$

Step 3: Evaluate the integral using the known integral of $\sec x$. Using the formula $\int \sec x \, dx = \ln \left| \tan \left( \frac{\pi}{4} + \frac{x}{2} \right) \right| + C$, we have $\frac{1}{\sqrt{2}} \int \sec \left(x + \frac{\pi}{4} \right) dx = \frac{1}{\sqrt{2}} \ln \left| \tan \left( \frac{\pi}{4} + \frac{x + \frac{\pi}{4}}{2} \right) \right| + C$

Step 4: Simplify the argument of the tangent function. We have $\frac{\pi}{4} + \frac{x + \frac{\pi}{4}}{2} = \frac{\pi}{4} + \frac{x}{2} + \frac{\pi}{8} = \frac{2\pi}{8} + \frac{\pi}{8} + \frac{x}{2} = \frac{3\pi}{8} + \frac{x}{2} = \frac{x}{2} + \frac{3\pi}{8}$ Thus, $\frac{1}{\sqrt{2}} \ln \left| \tan \left( \frac{\pi}{4} + \frac{x + \frac{\pi}{4}}{2} \right) \right| + C = \frac{1}{\sqrt{2}} \ln \left| \tan \left( \frac{x}{2} + \frac{3\pi}{8} \right) \right| + C$

Common Mistakes & Tips

• Remember the trigonometric transformation $a \cos x + b \sin x = R \cos(x - \alpha)$ and how to find $R$ and $\alpha$. Be careful with the sign of $\alpha$.
• The integral of $\sec x$ is $\ln |\sec x + \tan x| + C$ or $\ln \left| \tan \left( \frac{\pi}{4} + \frac{x}{2} \right) \right| + C$. Both are correct, but the latter is often more useful in these types of problems.
• Be careful with the algebraic manipulation of fractions within trigonometric functions.

Summary

We started by rewriting the integrand using the trigonometric identity $a \cos x + b \sin x = R \cos(x - \alpha)$. This allowed us to express the integral in terms of the secant function. We then used the known integral of the secant function and simplified the result to obtain the final answer.

Final Answer

The final answer is $\boxed{{1 \over {\sqrt 2 }}\log \left| {\tan \left( {{x \over 2} + {{3\pi } \over 8}} \right)} \right| + C}$, which corresponds to option (A).