int_0^{pi /4} frac{sec^2 x}{(1 + tan x)(2 + t | vedclass

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(D) Let $1 + 	an x = t$. Then,$\sec^2 x \,dx = dt$. When $x = 0$,$t = 1 + 	an(0) = 1$. When $x = \pi/4$,$t = 1 + 	an(\pi/4) = 2$. The integral beco

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$\log_e \left( \frac{4}{3} \right)$

Vedclass · Answer

(D) Let $1 + 	an x = t$. Then,$\sec^2 x \,dx = dt$. When $x = 0$,$t = 1 + 	an(0) = 1$. When $x = \pi/4$,$t = 1 + 	an(\pi/4) = 2$. The integral becomes $\int_1^2 \frac{dt}{t(t+1)}$. Using partial fractions,$\frac{1}{t(t+1)} = \frac{1}{t} - \frac{1}{t+1}$. Thus,$\int_1^2 \left( \frac{1}{t} - \frac{1}{t+1} ight) dt = [\log_e |t| - \log_e |t+1|]_1^2$. $= [\log_e |\frac{t}{t+1}|]_1^2 = \log_e \left( \frac{2}{3} ight) - \log_e \left( \frac{1}{2} ight) = \log_e \left( \frac{2/3}{1/2} ight) = \log_e \left( \frac{4}{3} ight)$.

$\int_0^{\pi /4} \frac{\sec^2 x}{(1 + \tan x)(2 + \tan x)} \,dx = $

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