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(A) Let $I = \int (\sqrt{	an x} + \sqrt{\cot x}) dx = \int \frac{	an x + 1}{\sqrt{	an x}} dx$. Substitute $	an x = t^2$,then $\sec^2 x dx = 2t dt$

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Let $I = \int (\sqrt{	an x} + \sqrt{\cot x}) dx = \int \frac{	an x + 1}{\sqrt{	an x}} dx$. Substitute $	an x = t^2$,then $\sec^2 x dx = 2t dt$. Since $\sec^2 x = 1 + 	an^2 x = 1 + t^4$,we have $dx = \frac{2t}{1+t^4} dt$. Thus,$I = \int \frac{t^2 + 1}{t} \cdot \frac{2t}{1+t^4} dt = 2 \int \frac{t^2 + 1}{t^4 + 1} dt$. 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 $t = \sqrt{	an x}$,we get $I = \sqrt{2} 	an^{-1}\left(\frac{	an x - 1}{\sqrt{2 	an x}}ight) + c$.

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

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