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:
We must simplify the integrand using trigonometric identities to remove the fraction before integrating using integration by parts.
Step 2: Key Formula or Approach:
Use the double angle identity: \(\cos 2x = \cos^2 x - \sin^2 x\).
Factor it as a difference of squares: \((\cos x - \sin x)(\cos x + \sin x)\) and cancel the denominator.
Use Integration by Parts: \(\int u \, dv = uv - \int v \, du\).
Step 3: Detailed Explanation:
Simplify the integrand:
\[ \frac{x(\cos^2 x - \sin^2 x)}{\cos x - \sin x} = \frac{x(\cos x - \sin x)(\cos x + \sin x)}{\cos x - \sin x} = x(\cos x + \sin x) \] Now evaluate the integral:
\[ I = \int x(\cos x + \sin x) dx \] Apply Integration by Parts.
Let \(u = x \implies du = dx\).
Let \(dv = (\cos x + \sin x)dx \implies v = \sin x - \cos x\).
\[ I = uv - \int v \, du \] \[ I = x(\sin x - \cos x) - \int (\sin x - \cos x) dx \] Integrate the remaining term:
\[ \int (\sin x - \cos x) dx = -\cos x - \sin x \] Substitute back:
\[ I = x(\sin x - \cos x) - (-\cos x - \sin x) + C \] \[ I = x(\sin x - \cos x) + \cos x + \sin x + C \] Step 4: Final Answer:
The result is \(x(\sin x - \cos x) + \cos x + \sin x + C\).