Question

Evaluate:

`∫ (1)/(sin^2 x cos^2 x) dx`

Answer 1

\[\int\frac{1}{\sin^2 x \cos^2 x}dx = 4\int\frac{1}{4 \sin^2 x \cos^2 x}dx\]

\[ = 4\int\frac{1}{\sin^2 \left( 2x \right)}dx\]

\[ = 4\int {cosec}^2 \left( 2x \right) dx\]

\[ = - \frac{4}{2}\cot\left( 2x \right) + c\]

\[ = - 2\cot\left( 2x \right) + c\]

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