int frac{sec^2 x}{log ((tan x)^{tan x})} dx = | vedclass

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Question

(C) Let $I = \int \frac{\sec^2 x}{\log ((	an x)^{	an x})} dx$. Using the property $\log(a^b) = b \log a$,the denominator becomes $	an x \cdot \log(

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Let $I = \int \frac{\sec^2 x}{\log ((	an x)^{	an x})} dx$. Using the property $\log(a^b) = b \log a$,the denominator becomes $	an x \cdot \log(	an x)$. So,$I = \int \frac{\sec^2 x}{	an x \cdot \log(	an x)} dx$. Let $	an x = t$,then $\sec^2 x dx = dt$. Substituting these into the integral,we get $I = \int \frac{dt}{t \log t}$. Now,let $\log t = z$,then $\frac{1}{t} dt = dz$. Substituting these,we get $I = \int \frac{dz}{z} = \log |z| + c$. Replacing $z$ with $\log t$,we get $I = \log |\log t| + c$. Finally,replacing $t$ with $	an x$,we get $I = \log |\log(	an x)| + c$.

$\int \frac{\sec^2 x}{\log ((\tan x)^{\tan x})} dx = $

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