tan-x-dx-

Question Number 134580 by metamorfose last updated on 05/Mar/21

$$\int\sqrt{{tan}\left({x}\right)}{dx}=…? \\ $$

Answered by Olaf last updated on 05/Mar/21

(1/( (√2)))arccos(cosx−sinx) +(1/( (√2)))ln(cosx+sinx+(√(sin2x)))+C

$$\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\mathrm{arccos}\left(\mathrm{cos}{x}−\mathrm{sin}{x}\right) \\ $$$$+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\mathrm{ln}\left(\mathrm{cos}{x}+\mathrm{sin}{x}+\sqrt{\mathrm{sin2}{x}}\right)+\mathrm{C} \\ $$

Answered by Ar Brandon last updated on 05/Mar/21

t^2 =tanx ⇒2tdt=(1+t^4 )dx I=∫t∙((2t)/(t^4 +1))dt=∫((2t^2 )/(t^4 +1))dt=∫{(((t^2 −1)+(t^2 +1))/(t^4 +1))}dt =∫ ((t^2 −1)/(t^4 +1))dt+∫ ((t^2 +1)/(t^4 +1))dt=∫((1−(1/t^2 ))/(t^2 +(1/t^2 )))dt+∫((1+(1/t^2 ))/(t^2 +(1/t^2 )))dt =∫((1−(1/t^2 ))/((t+(1/t))^2 −2))dt+∫((1+(1/t^2 ))/((t−(1/t))^2 +2))dt=∫(du/(u^2 −2))+∫(dv/(v^2 +2))

$$\mathrm{t}^{\mathrm{2}} =\mathrm{tanx}\:\Rightarrow\mathrm{2tdt}=\left(\mathrm{1}+\mathrm{t}^{\mathrm{4}} \right)\mathrm{dx} \\ $$$$\mathcal{I}=\int\mathrm{t}\centerdot\frac{\mathrm{2t}}{\mathrm{t}^{\mathrm{4}} +\mathrm{1}}\mathrm{dt}=\int\frac{\mathrm{2t}^{\mathrm{2}} }{\mathrm{t}^{\mathrm{4}} +\mathrm{1}}\mathrm{dt}=\int\left\{\frac{\left(\mathrm{t}^{\mathrm{2}} −\mathrm{1}\right)+\left(\mathrm{t}^{\mathrm{2}} +\mathrm{1}\right)}{\mathrm{t}^{\mathrm{4}} +\mathrm{1}}\right\}\mathrm{dt} \\ $$$$\:\:\:=\int\:\frac{\mathrm{t}^{\mathrm{2}} −\mathrm{1}}{\mathrm{t}^{\mathrm{4}} +\mathrm{1}}\mathrm{dt}+\int\:\frac{\mathrm{t}^{\mathrm{2}} +\mathrm{1}}{\mathrm{t}^{\mathrm{4}} +\mathrm{1}}\mathrm{dt}=\int\frac{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{t}^{\mathrm{2}} }}{\mathrm{t}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{t}^{\mathrm{2}} }}\mathrm{dt}+\int\frac{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{t}^{\mathrm{2}} }}{\mathrm{t}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{t}^{\mathrm{2}} }}\mathrm{dt} \\ $$$$\:\:\:=\int\frac{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{t}^{\mathrm{2}} }}{\left(\mathrm{t}+\frac{\mathrm{1}}{\mathrm{t}}\right)^{\mathrm{2}} −\mathrm{2}}\mathrm{dt}+\int\frac{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{t}^{\mathrm{2}} }}{\left(\mathrm{t}−\frac{\mathrm{1}}{\mathrm{t}}\right)^{\mathrm{2}} +\mathrm{2}}\mathrm{dt}=\int\frac{\mathrm{du}}{\mathrm{u}^{\mathrm{2}} −\mathrm{2}}+\int\frac{\mathrm{dv}}{\mathrm{v}^{\mathrm{2}} +\mathrm{2}} \\ $$

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