Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 151756 by Integrals last updated on 22/Aug/21

Commented by MJS_new last updated on 22/Aug/21

$$\int ({\cot }^{1/4}\beta +{\tan }^{1/4}\beta )d\beta =$$$$[t={\tan }^{1/4}\beta \to d\beta =\frac{4t^3}{t^8+1}dt]$$$$=4\int \frac{t^4+t^2}{t^8+1}dt=$$$$=\frac{\sqrt{2}}{2}\int (\frac{1}{t^2+\sqrt{2+\sqrt{2}}t+1}+\frac{1}{t^2-\sqrt{2+\sqrt{2}}t+1}-\frac{1}{t^2+\sqrt{2-\sqrt{2}}t+1}-\frac{1}{t^2-\sqrt{2-\sqrt{2}}t+1})dt$$$$\mathrm{these}\mathrm{can}\mathrm{be}\mathrm{solved}\mathrm{using}\mathrm{the}\mathrm{well}\mathrm{known}\mathrm{formula}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com