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

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Question

(C) We are given the integral $I = \int \frac{\sin ^{2} x-\cos ^{2} x}{\sin ^{2} x \cos ^{2} x} d x$.Split the fraction into two separate terms:$I = \

Vedclass · Accepted Answer

$	an x+\cot x+C$

Vedclass · Answer

(C) We are given the integral $I = \int \frac{\sin ^{2} x-\cos ^{2} x}{\sin ^{2} x \cos ^{2} x} d x$.Split the fraction into two separate terms:$I = \int \left( \frac{\sin ^{2} x}{\sin ^{2} x \cos ^{2} x} - \frac{\cos ^{2} x}{\sin ^{2} x \cos ^{2} x} ight) d x$.Simplify each term:$I = \int \left( \frac{1}{\cos ^{2} x} - \frac{1}{\sin ^{2} x} ight) d x$.Using trigonometric identities $\frac{1}{\cos ^{2} x} = \sec ^{2} x$ and $\frac{1}{\sin ^{2} x} = \operatorname{cosec} ^{2} x$:$I = \int (\sec ^{2} x - \operatorname{cosec} ^{2} x) d x$.Integrating term by term:$\int \sec ^{2} x d x = 	an x$ and $\int \operatorname{cosec} ^{2} x d x = -\cot x$.Therefore,$I = 	an x - (-\cot x) + C = 	an x + \cot x + C$.Hence,the correct option is $C$.

$\int \frac{\sin ^{2} x-\cos ^{2} x}{\sin ^{2} x \cos ^{2} x} d x$ is equal to

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