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Question Number 211708 by Skyneless last updated on 17/Sep/24

Answered by Frix last updated on 18/Sep/24

$$\int \sqrt{\tan x}dx=$$$$[t={2\tan }^{-1}\sqrt{\tan x}]$$$$=\int \frac{1-\cos t}{1+{\cos }^{2}t}dt$$$$$$$$\int \frac{1}{1+{\cos }^{2}t}dt=$$$$[u=\frac{\sqrt{2}\tan t}{2}]$$$$=\frac{\sqrt{2}}{2}\int \frac{du}{u^2+1}=\frac{\sqrt{2}{\tan }^{-1}u}{2}$$$$$$$$-\int \frac{\cos t}{1+{\cos }^{2}t}dt=$$$$[v=\frac{\sqrt{2}\sin t}{2}]$$$$=\frac{\sqrt{2}}{2}\int \frac{dv}{v^2-1}=-\frac{\sqrt{2}{\tanh }^{-1}v}{2}$$$$$$$$...\mathrm{which}\mathrm{is}\mathrm{nice}\mathrm{somehow}...$$

Answered by BHOOPENDRA last updated on 18/Sep/24

$$let\tan x=t^2$$$$\sec {}^{2}xdx=2tdt$$$$(1+t^4)dx=2tdt$$$$\Rightarrow \frac{2t}{(1+t^4)}dt=dx$$$$=\int \frac{2t^2}{(1+t^4)}dt$$$$=\int \frac{t^2+1}{1+t^4}dt+\int \frac{t^2-1}{1+t^4}dt$$$$=\int \frac{1+\frac{1}{t^2}}{\frac{1}{t^2}+t^2}dt+\int \frac{1-\frac{1}{t^2}}{\frac{1}{t^2}+t^2}dt\left(dividedby{t}^2\right)$$$$=\int \frac{1+\frac{1}{t^2}}{{(t-\frac{1}{t})}^2+2}dt+\int \frac{1-\frac{1}{t^2}}{{(t+\frac{1}{t})}^2-2}dt$$$$Let=t-1/t=p,t+1/t=q$$$$=1+\frac{1}{t^2}dt=dp,1-\frac{1}{t^2}dt=dq$$$$=\int \frac{1}{p^2+2}dp+\int \frac{1}{q^2-2}dq$$$$=\frac{1}{\sqrt{2}}{\tan }^{-1}\frac{p}{\sqrt{2}}+\frac{1}{2\sqrt{2}}log\left(\frac{q-\sqrt{2}}{q+\sqrt{2}}\right)+c$$$$substitutingallthevalues$$$$p=t-\frac{1}{t},q=t+\frac{1}{t}$$$$t=\sqrt{tanx}$$$$=\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\frac{\tan x-1}{\sqrt{2tanx}}\right)+\frac{1}{2\sqrt{2}}log\left(\frac{\tan x-\sqrt{2\tan x}+1}{\tan x+\sqrt{2\tan x}+1}\right)+c$$

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