int frac{sec^2 x , dx}{sqrt{tan^2 x + 4}} = | vedclass

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Question

(A) Let $t = 	an x$. Then,$dt = \sec^2 x \, dx$. Substituting these into the integral,we get: $\int \frac{dt}{\sqrt{t^2 + 2^2}}$. Using the standard

Vedclass · Accepted Answer

$\log \left[ 	an x + \sqrt{	an^2 x + 4} ight] + c$

Vedclass · Answer

(A) Let $t = 	an x$. Then,$dt = \sec^2 x \, dx$. Substituting these into the integral,we get: $\int \frac{dt}{\sqrt{t^2 + 2^2}}$. Using the standard integration formula $\int \frac{dx}{\sqrt{x^2 + a^2}} = \log |x + \sqrt{x^2 + a^2}| + c$,we obtain: $\log |t + \sqrt{t^2 + 2^2}| + c$. Substituting $t = 	an x$ back,the final result is: $\log |	an x + \sqrt{	an^2 x + 4}| + c$.

$\int \frac{\sec^2 x \, dx}{\sqrt{\tan^2 x + 4}} = $

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