int frac{dx}{sin^2 x cos^2 x} = | vedclass

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(C) To solve the integral $\int \frac{dx}{\sin^2 x \cos^2 x}$,we use the trigonometric identity $1 = \sin^2 x + \cos^2 x$.Substituting this into the n

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$	an x - \cot x + c$

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(C) To solve the integral $\int \frac{dx}{\sin^2 x \cos^2 x}$,we use the trigonometric identity $1 = \sin^2 x + \cos^2 x$.Substituting this into the numerator,we get:$\int \frac{\sin^2 x + \cos^2 x}{\sin^2 x \cos^2 x} dx$Splitting the fraction,we have:$\int \left( \frac{\sin^2 x}{\sin^2 x \cos^2 x} + \frac{\cos^2 x}{\sin^2 x \cos^2 x} ight) dx$This simplifies to:$\int (\sec^2 x + \csc^2 x) dx$Integrating term by term,we get:$\int \sec^2 x dx + \int \csc^2 x dx = 	an x - \cot x + c$.

$\int \frac{dx}{\sin^2 x \cos^2 x} = $

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