Question

\(\int e^{2x}\frac{2(sin2xcos2x-1)}{2sin^22x}\,dx = A\,e^{2x}cot2x+c\)
(Where c is the constant of integration.), then \(A^3 =\)

• A. \(\frac{1}{8}\)
• B. \(64\)
• C. \(8\)
• D. \(\frac{1}{64}\)

Hint:

Simplify the fraction to cot 2x - cosec^2 2x and use the result that e^(ax)[f(x) + f prime(x)/a] integrates to e^(ax) f(x)/a.

Answer 1

The Correct Option is A

Step 1: Simplify.
The integrand is $e^{2x}(\cot 2x - \csc^2 2x)$.

Step 2: Guess and differentiate.
Try $\dfrac{d}{dx}(e^{2x}\cot 2x) = 2e^{2x}\cot 2x + e^{2x}(-2\csc^2 2x) = 2e^{2x}(\cot 2x - \csc^2 2x)$.

Step 3: Compare.
The derivative of $e^{2x}\cot 2x$ is twice our integrand, so our integral is $\dfrac{1}{2}e^{2x}\cot 2x + c$.

Step 4: Result.
$A = 1/2$, so $A^3 = 1/8$.

Final Answer:
Option (A). \[ \boxed{\frac{1}{8}} \]