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Category: Integration

Question-151756

Question Number 151756 by Integrals last updated on 22/Aug/21 Commented by MJS_new last updated on 22/Aug/21 (cot1/4β+tan1/4β)dβ=$$\:\:\:\:\:\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] \