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MHTCET 13 Apr 2026 Shift 1 Paper

Question : 129 of 150

Marks: +1, -0

\int e^{2 x} \frac{2(\sin 2 x \cos 2 x-1)}{2 \sin ^{2} 2 x} d x=A e^{2 x} \cot 2 x+c

A^{3}=

Solution:

Concept:

The given integrand is a derivative of

e^{2x}\cot 2x

up to a constant factor.

Explanation:

Simplify the integrand by cancelling 2 from numerator and denominator:

\int e^{2x} \frac{\sin 2x \cos 2x - 1}{\sin^2 2x}\,dx

Split the fraction:

\frac{\sin 2x \cos 2x - 1}{\sin^2 2x} = \frac{\sin 2x \cos 2x}{\sin^2 2x} - \frac{1}{\sin^2 2x} = \cot 2x - \csc^2 2x

So the integrand becomes

e^{2x}(\cot 2x - \csc^2 2x)

Differentiate

e^{2x}\cot 2x

\frac{d}{dx}\left(e^{2x}\cot 2x\right) = 2e^{2x}\cot 2x - 2e^{2x}\csc^2 2x = 2e^{2x}(\cot 2x - \csc^2 2x)

Therefore,

e^{2x}(\cot 2x - \csc^2 2x) = \frac{1}{2}\frac{d}{dx}\left(e^{2x}\cot 2x\right)

Integrating both sides:

\int e^{2x}(\cot 2x - \csc^2 2x)\,dx = \frac{1}{2}e^{2x}\cot 2x + c

Comparing with the given form

Ae^{2x}\cot 2x + c

, we get

A = \frac{1}{2}

Thus,

A^3 = \left(\frac{1}{2}\right)^3 = \frac{1}{8}

Answer:

A^3 = \frac{1}{8}

Option A.

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