int (sqrt{tan x} + sqrt{cot x}) , dx = | vedclass

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(C) Let $I = \int (\sqrt{	an x} + \sqrt{\cot x}) \, dx = \int \left( \sqrt{	an x} + \frac{1}{\sqrt{	an x}} ight) \, dx = \int \frac{	an x + 1}{\

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$\sqrt{2} 	an^{-1} \left( \frac{	an x - 1}{\sqrt{2 	an x}} ight) + c$

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(C) Let $I = \int (\sqrt{	an x} + \sqrt{\cot x}) \, dx = \int \left( \sqrt{	an x} + \frac{1}{\sqrt{	an x}} ight) \, dx = \int \frac{	an x + 1}{\sqrt{	an x}} \, dx$. Substitute $\sqrt{	an x} = t$,so $	an x = t^2$ and $\sec^2 x \, dx = 2t \, dt$. Since $\sec^2 x = 1 + 	an^2 x = 1 + t^4$,we have $dx = \frac{2t}{1 + t^4} \, dt$. Substituting these into the integral: $I = \int \frac{t^2 + 1}{t} \cdot \frac{2t}{1 + t^4} \, dt = 2 \int \frac{t^2 + 1}{t^4 + 1} \, dt$. Divide numerator and denominator by $t^2$: $I = 2 \int \frac{1 + 1/t^2}{t^2 + 1/t^2} \, dt = 2 \int \frac{1 + 1/t^2}{(t - 1/t)^2 + 2} \, dt$. Let $u = t - 1/t$,then $du = (1 + 1/t^2) \, dt$. $I = 2 \int \frac{du}{u^2 + (\sqrt{2})^2} = 2 \cdot \frac{1}{\sqrt{2}} 	an^{-1} \left( \frac{u}{\sqrt{2}} ight) + c = \sqrt{2} 	an^{-1} \left( \frac{t - 1/t}{\sqrt{2}} ight) + c$. Substituting $t = \sqrt{	an x}$: $I = \sqrt{2} 	an^{-1} \left( \frac{	an x - 1}{\sqrt{2 	an x}} ight) + c$.

$\int (\sqrt{\tan x} + \sqrt{\cot x}) \, dx = $

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