MHT CET · Maths · Indefinite Integration
\(\int \mathrm{e}^{2 x} \frac{(\sin 2 x \cos 2 x-1)}{\sin ^2 2 x} \mathrm{~d} x=\)
- A \(\mathrm{e}^{2 x} \cot (2 x)+\mathrm{c}\), where c is the constant of integration
- B \(2 \mathrm{e}^{2 x} \cot (2 x)+\mathrm{c}\), where c is the constant of integration
- C \(4 e^{2 x} \cot (2 x)+c\), where \(c\) is the constant of integration
- D \(\frac{1}{2} \mathrm{e}^{2 x} \cot (2 x)+\mathrm{c}\), where c is the constant of integration
Answer & Solution
Correct Answer
(D) \(\frac{1}{2} \mathrm{e}^{2 x} \cot (2 x)+\mathrm{c}\), where c is the constant of integration
Step-by-step Solution
Detailed explanation
\( \int \mathrm{e}^{2 x} \frac{(\sin 2 x \cos 2 x-1)}{\sin ^2 2 x} \mathrm{~d} x = \int \mathrm{e}^{2 x} \left(\frac{\cos 2 x}{\sin 2 x} - \frac{1}{\sin ^2 2 x}\right) \mathrm{~d} x \) \( = \int \mathrm{e}^{2 x} (\cot 2 x - \csc ^2 2 x) \mathrm{~d} x \)
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