int {{ 1 + 2tan x(tan x + sec x)} }^{1/2} dx | vedclass

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Question

(C) Given integral is $I = \int {\{ 1 + 2	an^2 x + 2	an x \sec x \}^{1/2}} dx$. Since $1 + 	an^2 x = \sec^2 x$,we can rewrite the expression inside

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Given integral is $I = \int {\{ 1 + 2	an^2 x + 2	an x \sec x \}^{1/2}} dx$. Since $1 + 	an^2 x = \sec^2 x$,we can rewrite the expression inside the square root as: $1 + 2	an^2 x + 2	an x \sec x = \sec^2 x + 	an^2 x + 2	an x \sec x = (\sec x + 	an x)^2$. Substituting this back into the integral: $I = \int (\sec x + 	an x) dx$. 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 $\log a + \log b = \log(ab)$: $I = \log |\sec x(\sec x + 	an x)| + c$. Thus,the correct option is $C$.

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

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