int (sec x + tan x)^2 dx = | vedclass

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(A) We need to evaluate the integral $I = \int (\sec x + 	an x)^2 dx$. Expanding the integrand using the identity $(a+b)^2 = a^2 + b^2 + 2ab$: $I = \

Vedclass · Accepted Answer

$2(\sec x + 	an x) - x + c$

Vedclass · Answer

(A) We need to evaluate the integral $I = \int (\sec x + 	an x)^2 dx$. Expanding the integrand using the identity $(a+b)^2 = a^2 + b^2 + 2ab$: $I = \int (\sec^2 x + 	an^2 x + 2\sec x 	an x) dx$. Using the trigonometric identity $	an^2 x = \sec^2 x - 1$: $I = \int (\sec^2 x + (\sec^2 x - 1) + 2\sec x 	an x) dx$. $I = \int (2\sec^2 x - 1 + 2\sec x 	an x) dx$. Integrating term by term: $I = 2 \int \sec^2 x dx - \int 1 dx + 2 \int \sec x 	an x dx$. $I = 2	an x - x + 2\sec x + c$. $I = 2(\sec x + 	an x) - x + c$.

$\int (\sec x + \tan x)^2 dx = $

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