Question

\(\int e^{x}\left(\log (\sin x)+\sqrt{\frac{1+\cos 2x}{1-\cos 2x}}\right)dx=\ldots\)

• A. \(e^{x}\cot x+C\)
• B. \(e^{x}\log \sin x+C\)
• C. \(e^{x}\log \sin x+\tan x+C\)
• D. \(e^{x}+\sin x+C\)

Hint:

Simplify the square root to cot x, then use the identity that the integral of e^x (f(x) + f'(x)) equals e^x f(x).

Answer 1

The Correct Option is B

Step 1: Simplify the root
\(\sqrt{\frac{1+\cos 2x}{1-\cos 2x}}=\sqrt{\frac{2\cos ^{2}x}{2\sin ^{2}x}}=\cot x\).

Step 2: Identify f and f'
Take \(f(x)=\log \sin x\). Then \(f'(x)=\cot x\).

Step 3: Apply the formula
\(\int e^{x}[f(x)+f'(x)]dx=e^{x}f(x)+C\).

Step 4: Write the result
The integral equals \(e^{x}\log \sin x+C\).

Final Answer:
The integral equals \(e^{x}\log \sin x+C\). \[ \boxed{\text{(B) }e^{x}\log \sin x+C} \]