Question

\( \int \left(\frac{2 - \sin 2x}{1 - \cos 2x}\right) e^{x} \, dx \) is equal to

• A. $-e^{x} \cot x + c$
• B. $e^{x} \cot x + c$
• C. $2e^{x} \cot x + c$
• D. $-2e^{x} \cot x + c$

Hint:

Standard $e^x$ form: $\int e^x[f(x) + f'(x)] dx = e^x f(x)$.

Answer 1

The Correct Option is A

Step 1: Trigonometric Simplification
$I = \int \left(\frac{2 - 2 \sin x \cos x}{2 \sin^{2}x}\right)e^{x} dx = \int (\csc^{2}x - \cot x)e^{x} dx$.
Step 2: Apply Integration Property
Use the rule $\int e^{x}[f(x) + f'(x)] dx = e^{x}f(x) + c$. Here, $f(x) = -\cot x$ and $f'(x) = \csc^{2}x$.
Step 3: Conclusion
Result $= e^{x}(-\cot x) + c = -e^{x} \cot x + c$.
Final Answer: (a)