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Question Number 40830 by math khazana by abdo last updated on 28/Jul/18

$$find\int \sqrt{2+{tan}^{2}t}dt$$

Commented by maxmathsup by imad last updated on 31/Jul/18

$$letI=\int \sqrt{2+{tan}^{2}t}dtchangementtant=\sqrt{2}sh\left(x\right)\Rightarrow t=arctan\left(\sqrt{2}shx\right)$$$$\Rightarrow dt=\frac{\sqrt{2}ch\left(x\right)}{1+2{sh}^{2}x}dx$$$$I=\int \sqrt{2+2{sh}^{2}x}\frac{\sqrt{2}ch\left(x\right)}{1+2{sh}^{2}x}dx=\int \frac{2{ch}^{2}x}{1+2{sh}^2}dx$$$$=\int \frac{1+ch\left(2x\right)}{1+ch\left(2x\right)-1}dx=\int \frac{ch\left(2x\right)+1}{ch\left(2x\right)}dx=x+\int \frac{dx}{ch\left(2x\right)}but$$$$\int \frac{dx}{ch\left(2x\right)}=\int \frac{2dx}{e^{2x}+e^{-2x}}{=}_{e^x=t}\int \frac{2}{t^2+t^{-2}}\frac{dt}{t}$$$$=\int \frac{2dt}{t^3+\frac{1}{t}}=\int \frac{2tdt}{t^4+1}{=}_{t^2=u}\int \frac{du}{1+u^2}=arctan\left(u\right)+c$$$$=arctan\left(t^2\right)+c=actan\left(e^{2x}\right)+cbutx=argsh\left(\frac{tant}{\sqrt{2}}\right)$$$$=ln(\frac{tant}{\sqrt{2}}+\sqrt{1+\frac{{tan}^{2}t}{2})})$$$$I=argsh\left(\frac{tant}{\sqrt{2}}\right)+arctan\left({\{\frac{tant}{\sqrt{2}}+\sqrt{1+\frac{{tan}^{2}t}{2}}\}}^2\right)+c$$

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