int [1+2 tan x(tan x+sec x)]^{frac{1}{2}} dx | vedclass

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Question

(C) Let $I = \int [1+2 	an^2 x + 2 	an x \sec x]^{1/2} dx$. Since $1 + 	an^2 x = \sec^2 x$,we can rewrite the expression inside the square root as:

Vedclass · Accepted Answer

$\log [\sec x(\sec x+	an x)]+c$

Vedclass · Answer

(C) Let $I = \int [1+2 	an^2 x + 2 	an x \sec x]^{1/2} dx$. Since $1 + 	an^2 x = \sec^2 x$,we can rewrite the expression inside the square root as: $I = \int [\sec^2 x + 	an^2 x + 2 \sec x 	an x]^{1/2} dx$. This is a perfect square: $I = \int [(\sec x + 	an x)^2]^{1/2} dx = \int (\sec x + 	an x) dx$. Integrating term by term: $I = \int \sec x dx + \int 	an x dx$. $I = \log |\sec x + 	an x| + \log |\sec x| + c$. Using the property $\log a + \log b = \log(ab)$: $I = \log |\sec x(\sec x + 	an x)| + c$.

$\int [1+2 \tan x(\tan x+\sec x)]^{\frac{1}{2}} dx = $

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