int frac{sec x}{sqrt{log (sec x+tan x)}} d x= | vedclass

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Question

(D) Let $I = \int \frac{\sec x}{\sqrt{\log (\sec x+	an x)}} dx$. Substitute $t = \log (\sec x + 	an x)$. Differentiating both sides with respect to

Vedclass · Accepted Answer

$2 \sqrt{\log (\sec x+	an x)}+c$

Vedclass · Answer

(D) Let $I = \int \frac{\sec x}{\sqrt{\log (\sec x+	an x)}} dx$. Substitute $t = \log (\sec x + 	an x)$. Differentiating both sides with respect to $x$,we get: $dt = \frac{1}{\sec x + 	an x} (\sec x 	an x + \sec^2 x) dx$. $dt = \frac{\sec x(	an x + \sec x)}{\sec x + 	an x} dx$. $dt = \sec x dx$. Now,the integral becomes: $I = \int \frac{dt}{\sqrt{t}} = \int t^{-1/2} dt$. Integrating with respect to $t$: $I = \frac{t^{1/2}}{1/2} + c = 2\sqrt{t} + c$. Substituting back $t = \log (\sec x + 	an x)$: $I = 2\sqrt{\log (\sec x + 	an x)} + c$.

$\int \frac{\sec x}{\sqrt{\log (\sec x+\tan x)}} d x=$

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