Question Number 224714 by fantastic last updated on 28/Sep/25
$$\int\frac{\mathrm{1}}{\:\sqrt{\mathrm{tan}\:\theta}}\:{d}\theta \\ $$
Commented by Ghisom_ last updated on 29/Sep/25
![∫(dθ/( (√(tan θ))))= [t=(√(tan θ)) → dθ=2cos^2 θ (√(tan θ))] =2∫(dt/(t^4 +1))=Σ_(j=1) ^4 I_j I_1 =(1/(2(√2)))∫((2t+(√2))/(t^2 +(√2)t+1))dt=((√2)/4)ln (t^2 +(√2)t+1) I_2 =(1/2)∫(dt/(t^2 +(√2)t+1))=((√2)/2)arctan ((√2)t+1) I_3 =−(1/(2(√2)))∫((2t−(√2))/(t^2 −(√2)t+1))dt=−((√2)/4)ln (t^2 −(√2)t+1) I_4 =(1/2)∫(dt/(t^2 −(√2)t+1))=((√2)/2)arctan ((√2)t−1) the rest is easy](https://www.tinkutara.com/question/Q224716.png)
$$\int\frac{{d}\theta}{\:\sqrt{\mathrm{tan}\:\theta}}= \\ $$$$\:\:\:\:\:\left[{t}=\sqrt{\mathrm{tan}\:\theta}\:\rightarrow\:{d}\theta=\mathrm{2cos}^{\mathrm{2}} \:\theta\:\sqrt{\mathrm{tan}\:\theta}\right] \\ $$$$=\mathrm{2}\int\frac{{dt}}{{t}^{\mathrm{4}} +\mathrm{1}}=\underset{{j}=\mathrm{1}} {\overset{\mathrm{4}} {\sum}}{I}_{{j}} \\ $$$${I}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\int\frac{\mathrm{2}{t}+\sqrt{\mathrm{2}}}{{t}^{\mathrm{2}} +\sqrt{\mathrm{2}}{t}+\mathrm{1}}{dt}=\frac{\sqrt{\mathrm{2}}}{\mathrm{4}}\mathrm{ln}\:\left({t}^{\mathrm{2}} +\sqrt{\mathrm{2}}{t}+\mathrm{1}\right) \\ $$$${I}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dt}}{{t}^{\mathrm{2}} +\sqrt{\mathrm{2}}{t}+\mathrm{1}}=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\mathrm{arctan}\:\left(\sqrt{\mathrm{2}}{t}+\mathrm{1}\right) \\ $$$${I}_{\mathrm{3}} =−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\int\frac{\mathrm{2}{t}−\sqrt{\mathrm{2}}}{{t}^{\mathrm{2}} −\sqrt{\mathrm{2}}{t}+\mathrm{1}}{dt}=−\frac{\sqrt{\mathrm{2}}}{\mathrm{4}}\mathrm{ln}\:\left({t}^{\mathrm{2}} −\sqrt{\mathrm{2}}{t}+\mathrm{1}\right) \\ $$$${I}_{\mathrm{4}} =\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dt}}{{t}^{\mathrm{2}} −\sqrt{\mathrm{2}}{t}+\mathrm{1}}=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\mathrm{arctan}\:\left(\sqrt{\mathrm{2}}{t}−\mathrm{1}\right) \\ $$$$\mathrm{the}\:\mathrm{rest}\:\mathrm{is}\:\mathrm{easy} \\ $$