e² x} {( 2 x 2 x-1)}{ ² 2 x} d x=

Correct Answer is : 12e2xcot⁡(2x)+c1/2 e² x} (2 x)+c21e2xcot(2x)+c, where c is the constant of integration

e² x} (2 x)+c, where c is the constant of integration

2 e² x} (2 x)+c, where c is the constant of integration

4 e² x} (2 x)+c, where c is the constant of integration

1/2 e² x} (2 x)+c, where c is the constant of integration

Correct Answer is : 12e2xcot⁡(2x)+c1/2 e² x} (2 x)+c21e2xcot(2x)+c, where c is the constant of integration

Test Index

MHTCET 22 Apr 2025 Shift 2 Paper

Question : 142 of 150

Marks: +1, -0

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

Solution:

Concept:

The integral is of the form

\int e^{x}[f(x)+f'(x)]\,dx = e^{x}f(x) + c

We simplify the given expression and use a suitable substitution.

Explanation:

First, simplify the integrand by dividing each term in the numerator by

\sin^2 2x

e^{2x}\left(\frac{\sin 2x \cos 2x - 1}{\sin^2 2x}\right) = e^{2x}\left(\frac{\sin 2x \cos 2x}{\sin^2 2x} - \frac{1}{\sin^2 2x}\right)

This becomes:

e^{2x}(\cot 2x - \mathrm{cosec}^2\,2x)

Put

t = 2x

, so

dx = \frac{dt}{2}

Then the integral becomes:

I = \frac{1}{2}\int e^{t}(\cot t - \mathrm{cosec}^2 t)\,dt

Now,

\frac{d}{dt}(\cot t) = -\mathrm{cosec}^2 t

Hence

\cot t - \mathrm{cosec}^2 t = \cot t + \frac{d}{dt}(\cot t)

Using the formula

\int e^{t}[f(t) + f'(t)]\,dt = e^{t}f(t) + c

, we get:

I = \frac{1}{2}e^{t}\cot t + c

Substituting back

t = 2x

I = \frac{1}{2}e^{2x}\cot 2x + c

Answer:

\frac{1}{2}e^{2x}\cot(2x) + c

Hence, the correct option is D.

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