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

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Question

(C) Let $1+	an x = t$. Then $\sec^2 x \, dx = dt$. When $x = 0$,$t = 1+	an(0) = 1$. When $x = \frac{\pi}{4}$,$t = 1+	an(\frac{\pi}{4}) = 2$. The in

Vedclass · Accepted Answer

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

Vedclass · Answer

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

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

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