Question

$\int \frac{x \cos 2x}{\cos x - \sin x} dx = $}

• A. $x(\sin x - \cos x) + \cos x + \sin x + C$
• B. $x(\cos x - \sin x) + C$
• C. $x(\sin x + \cos x) + \sin x - \cos x + C$
• D. $x(\sin x + \cos x) - \cos x \sin x + C$
• E. $x \cos x \sin x + C$

Hint:

Whenever a trigonometric fraction has a double angle in the numerator, check if factoring it will cancel the denominator. This usually turns a hard rational integral into a basic integration by parts problem.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
First, simplify the integrand using trigonometric identities. Then, use integration by parts to solve the resulting integral.
Key Formula or Approach:
1. Use \( \cos 2x = \cos^2 x - \sin^2 x = (\cos x - \sin x)(\cos x + \sin x) \).
2. Integration by Parts: \( \int u dv = uv - \int v du \).

Step 2: Detailed Explanation:
Simplify the integral:
\[ I = \int \frac{x(\cos x - \sin x)(\cos x + \sin x)}{\cos x - \sin x} dx = \int x(\cos x + \sin x) dx \]
Apply Integration by Parts:
Let \( u = x \implies du = dx \).
Let \( dv = (\cos x + \sin x) dx \implies v = \sin x - \cos x \).
Using the formula:
\[ I = x(\sin x - \cos x) - \int (\sin x - \cos x) dx \]
\[ I = x(\sin x - \cos x) - [-\cos x - \sin x] + C \]
\[ I = x(\sin x - \cos x) + \cos x + \sin x + C \]

Step 3: Final Answer:
The integral is \( x(\sin x - \cos x) + \cos x + \sin x + C \).