int_0^{pi /4} sec x log (sec x + tan x) , dx | vedclass

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Question

(A) Let $I = \int_0^{\pi /4} \sec x \log (\sec x + 	an x) \, dx$. Substitute $t = \log (\sec x + 	an x)$. Then $dt = \frac{1}{\sec x + 	an x} (\sec

Vedclass · Accepted Answer

$\frac{1}{2} [\log (1 + \sqrt{2})]^2$

Vedclass · Answer

(A) Let $I = \int_0^{\pi /4} \sec x \log (\sec x + 	an x) \, dx$. Substitute $t = \log (\sec x + 	an x)$. Then $dt = \frac{1}{\sec x + 	an x} (\sec x 	an x + \sec^2 x) \, dx = \frac{\sec x (	an x + \sec x)}{\sec x + 	an x} \, dx = \sec x \, dx$. Change the limits of integration: When $x = 0$,$t = \log (\sec 0 + 	an 0) = \log (1 + 0) = \log 1 = 0$. When $x = \pi / 4$,$t = \log (\sec(\pi / 4) + 	an(\pi / 4)) = \log (\sqrt{2} + 1)$. Now,the integral becomes: $I = \int_0^{\log (\sqrt{2} + 1)} t \, dt = \left[ \frac{t^2}{2} ight]_0^{\log (\sqrt{2} + 1)} = \frac{1}{2} [\log (\sqrt{2} + 1)]^2$. Thus,the correct option is $A$.

$\int_0^{\pi /4} \sec x \log (\sec x + \tan x) \, dx = $

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