int sec^2 x csc^2 x , dx = . . . . . . + C | vedclass

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(A) We have the integral $I = \int \sec^2 x \csc^2 x \, dx$.Using the identities $\sec^2 x = \frac{1}{\cos^2 x}$ and $\csc^2 x = \frac{1}{\sin^2 x}$,

Vedclass · Accepted Answer

$	an x - \cot x$

Vedclass · Answer

(A) We have the integral $I = \int \sec^2 x \csc^2 x \, dx$.Using the identities $\sec^2 x = \frac{1}{\cos^2 x}$ and $\csc^2 x = \frac{1}{\sin^2 x}$, we get:$I = \int \frac{1}{\cos^2 x \sin^2 x} \, dx$.Since $1 = \sin^2 x + \cos^2 x$, we can write:$I = \int \frac{\sin^2 x + \cos^2 x}{\cos^2 x \sin^2 x} \, dx$.$I = \int \left( \frac{\sin^2 x}{\cos^2 x \sin^2 x} + \frac{\cos^2 x}{\cos^2 x \sin^2 x} ight) \, dx$.$I = \int (\sec^2 x + \csc^2 x) \, dx$.Integrating term by term, we get:$I = 	an x - \cot x + C$.

$\int \sec^2 x \csc^2 x \, dx = $ . . . . . . $+ C$

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