int frac{sin^2 x - cos^2 x}{sin^2 x cos^2 x} | vedclass

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Question

(A) Given integral: $I = \int \frac{\sin^2 x - \cos^2 x}{\sin^2 x \cos^2 x} dx$ Split the fraction: $I = \int \left( \frac{\sin^2 x}{\sin^2 x \cos^2 x

Vedclass · Accepted Answer

$	an x + \cot x + c$

Vedclass · Answer

(A) Given integral: $I = \int \frac{\sin^2 x - \cos^2 x}{\sin^2 x \cos^2 x} dx$ Split the fraction: $I = \int \left( \frac{\sin^2 x}{\sin^2 x \cos^2 x} - \frac{\cos^2 x}{\sin^2 x \cos^2 x} ight) dx$ Simplify the terms: $I = \int \left( \frac{1}{\cos^2 x} - \frac{1}{\sin^2 x} ight) dx$ Use trigonometric identities: $I = \int (\sec^2 x - \csc^2 x) dx$ Integrate: $I = 	an x - (-\cot x) + c$ Final result: $I = 	an x + \cot x + c$

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

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