int sec x , dx = | vedclass

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Question

(B) The integral of $\sec x$ is given by: $\int \sec x \, dx = \log |\sec x + 	an x| + c$ We know that $\sec x + 	an x = \frac{1}{\sec x - 	an x}$.

Vedclass · Accepted Answer

$-\log (\sec x - 	an x) + c$

Vedclass · Answer

(B) The integral of $\sec x$ is given by: $\int \sec x \, dx = \log |\sec x + 	an x| + c$ We know that $\sec x + 	an x = \frac{1}{\sec x - 	an x}$. Substituting this into the expression: $\int \sec x \, dx = \log \left| \frac{1}{\sec x - 	an x} ight| + c$ Using the property $\log(a^{-1}) = -\log(a)$: $\int \sec x \, dx = -\log |\sec x - 	an x| + c$. Thus,option $B$ is correct.

$\int \sec x \, dx = $

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