Question Number 151756 by Integrals last updated on 22/Aug/21
Commented by MJS_new last updated on 22/Aug/21
![∫(cot^(1/4) β +tan^(1/4) β)dβ= [t=tan^(1/4) β → dβ=((4t^3 )/(t^8 +1))dt] =4∫((t^4 +t^2 )/(t^8 +1))dt= =((√2)/2)∫((1/(t^2 +(√(2+(√2)))t+1))+(1/(t^2 −(√(2+(√2)))t+1))−(1/(t^2 +(√(2−(√2)))t+1))−(1/(t^2 −(√(2−(√2)))t+1)))dt these can be solved using the well known formula](Q151769.png)
$$\int\left(\mathrm{cot}^{\mathrm{1}/\mathrm{4}} \:\beta\:+\mathrm{tan}^{\mathrm{1}/\mathrm{4}} \:\beta\right){d}\beta= \\ $$$$\:\:\:\:\:\left[{t}=\mathrm{tan}^{\mathrm{1}/\mathrm{4}} \:\beta\:\rightarrow\:{d}\beta=\frac{\mathrm{4}{t}^{\mathrm{3}} }{{t}^{\mathrm{8}} +\mathrm{1}}{dt}\right] \\ $$$$=\mathrm{4}\int\frac{{t}^{\mathrm{4}} +{t}^{\mathrm{2}} }{{t}^{\mathrm{8}} +\mathrm{1}}{dt}= \\ $$$$=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\int\left(\frac{\mathrm{1}}{{t}^{\mathrm{2}} +\sqrt{\mathrm{2}+\sqrt{\mathrm{2}}}{t}+\mathrm{1}}+\frac{\mathrm{1}}{{t}^{\mathrm{2}} −\sqrt{\mathrm{2}+\sqrt{\mathrm{2}}}{t}+\mathrm{1}}−\frac{\mathrm{1}}{{t}^{\mathrm{2}} +\sqrt{\mathrm{2}−\sqrt{\mathrm{2}}}{t}+\mathrm{1}}−\frac{\mathrm{1}}{{t}^{\mathrm{2}} −\sqrt{\mathrm{2}−\sqrt{\mathrm{2}}}{t}+\mathrm{1}}\right){dt} \\ $$$$\mathrm{these}\:\mathrm{can}\:\mathrm{be}\:\mathrm{solved}\:\mathrm{using}\:\mathrm{the}\:\mathrm{well}\:\mathrm{known}\:\mathrm{formula} \\ $$