Question-151756

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} β + \tan^{1 / 4} β) d β =

[t = \tan^{1 / 4} β \to d β = \frac{4 t^{3}}{t^{8} + 1} d t]

= 4 \int \frac{t^{4} + t^{2}}{t^{8} + 1} d t =

= \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}) d t

these can be solved using the well known formula

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