Question

If
\[ \int \left(\frac{2-\sin2x}{1-\cos2x}\right)e^x\,dx \] 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:

Try matching integrand with derivative of a product \(e^x\cdot(\text{trig})\). Product rule often directly gives the answer.

Answer 1

The Correct Option is A

Step 1: Simplify trigonometric part.
\[ \frac{2-\sin2x}{1-\cos2x} \]
Use identities:
\[ 1-\cos2x=2\sin^2x \]
\[ \sin2x=2\sin x\cos x \]
So:
\[ \frac{2-2\sin x\cos x}{2\sin^2x} =\frac{1-\sin x\cos x}{\sin^2x} \]
\[ =\csc^2x-\cot x \]
Step 2: Integral becomes.
\[ \int (\csc^2x-\cot x)e^x dx \]
Step 3: Observe derivative form.
Differentiate \(-e^x\cot x\):
\[ \frac{d}{dx}(-e^x\cot x) =-e^x\cot x + e^x\csc^2x =e^x(\csc^2x-\cot x) \]
So integral equals:
\[ -e^x\cot x + c \]
Final Answer:
\[ \boxed{-e^x\cot x + c} \]