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: Understanding the Concept:
The integrand has the form \(e^{2x}\,[\,\text{something}\,]\). We try to write the bracket as \(g(x) + \dfrac{g'(x)}{2}\) with \(g = \cot 2x\).

Step 2: Simplify the fraction.
\[ \frac{2(\sin 2x\cos 2x - 1)}{2\sin^2 2x} = \frac{\sin 2x\cos 2x - 1}{\sin^2 2x} = \cot 2x - \operatorname{cosec}^2 2x \]

Step 3: Check the form.
Let \(g(x) = \cot 2x\). Then \(g'(x) = -2\operatorname{cosec}^2 2x\), so \(\dfrac{g'(x)}{2} = -\operatorname{cosec}^2 2x\). The bracket is exactly \(g + \dfrac{g'}{2}\).

Step 4: Integrate.
Since \(\dfrac{d}{dx}\left[e^{2x}g\right] = e^{2x}(2g + g') = 2e^{2x}\left(g + \dfrac{g'}{2}\right)\), we get
\[ \int e^{2x}\left(g + \frac{g'}{2}\right)dx = \frac{1}{2}e^{2x}\cot 2x + c \]

Step 5: Find \(A^3\).
So \(A = \dfrac{1}{2}\) and \(A^3 = \dfrac{1}{8}\).

Final Answer:
\(A^3 = \dfrac{1}{8}\), option (A). \[ \boxed{\frac{1}{8}} \]