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

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Question

(B) We have the integral $I = \int \frac{dx}{\sin^2 x \cos^2 x}$.Since $1 = \sin^2 x + \cos^2 x$,we can write the integral as:$I = \int \frac{\sin^2 x

Vedclass · Accepted Answer

$	an x - \cot x + c$,where $c$ is the constant of integration.

Vedclass · Answer

(B) We have the integral $I = \int \frac{dx}{\sin^2 x \cos^2 x}$.Since $1 = \sin^2 x + \cos^2 x$,we can write the integral as:$I = \int \frac{\sin^2 x + \cos^2 x}{\sin^2 x \cos^2 x} dx$$I = \int \left( \frac{\sin^2 x}{\sin^2 x \cos^2 x} + \frac{\cos^2 x}{\sin^2 x \cos^2 x} ight) dx$$I = \int \left( \frac{1}{\cos^2 x} + \frac{1}{\sin^2 x} ight) dx$$I = \int (\sec^2 x + \csc^2 x) dx$Using the standard integrals $\int \sec^2 x dx = 	an x$ and $\int \csc^2 x dx = -\cot x$,we get:$I = 	an x - \cot x + c$.

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

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