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प्रश्न
Evaluate:
उत्तर
\[ = 4\int\frac{1}{\sin^2 \left( 2x \right)}dx\]
\[ = 4\int {cosec}^2 \left( 2x \right) dx\]
\[ = - \frac{4}{2}\cot\left( 2x \right) + c\]
\[ = - 2\cot\left( 2x \right) + c\]
\[\text{ Hence,} \int\frac{1}{\sin^2 x \cos^2 x}dx = - \text{ 2 cot}\left( \text{ 2x }\right) + c\]
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