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Category: Integration

Question-193526

Question Number 193526 by SaRahAli last updated on 16/Jun/23 Answered by MM42 last updated on 16/Jun/23 $${I}=\int\:\frac{{sinx}×{cosx}}{{sinx}+{cos}^{\mathrm{2}} {x}}\:{dx}=−\int\:\frac{{sinx}×{cosx}}{{sinx}−{sinx}−\mathrm{1}}\:{dx} \\ $$$${let}\:\:\:{sinx}={u}\Rightarrow{cosxdx}={du} \\ $$$$\Rightarrow{I}=−\int\:\frac{{udu}}{{u}^{\mathrm{2}} −{u}−\mathrm{1}}\:{du}\:=\int\:\frac{{A}}{{u}−{u}_{\mathrm{1}} }\:{du}\:+\int\:\frac{{B}}{{u}−{u}_{\mathrm{2}} }\:{du}\:…

dx-x-6-1-

Question Number 193538 by cortano12 last updated on 16/Jun/23 $$\:\:\:\int\:\frac{\mathrm{dx}}{\mathrm{x}^{\mathrm{6}} +\mathrm{1}}\:=? \\ $$ Answered by Subhi last updated on 16/Jun/23 $$\int\frac{\mathrm{1}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)\left({x}^{\mathrm{4}} −{x}^{\mathrm{2}} +\mathrm{1}\right)}\:\Rrightarrow\:\int\frac{{a}}{{x}^{\mathrm{2}} +\mathrm{1}}+\frac{{bx}^{\mathrm{2}}…

Question-193512

Question Number 193512 by SaRahAli last updated on 15/Jun/23 Answered by Subhi last updated on 15/Jun/23 $$\int\frac{\mathrm{1}}{\frac{\mathrm{1}}{{cos}\left({x}\right)}.\frac{{sin}\left({x}\right)}{{cos}\left({x}\right)}}\:\Rrightarrow\:\int\frac{{cos}^{\mathrm{2}} \left({x}\right)}{{sin}\left({x}\right)}.{dx} \\ $$$$\int\frac{\mathrm{1}−{sin}^{\mathrm{2}} \left({x}\right)}{{sin}\left({x}\right)}.{dx}\:\:\looparrowright\:\int{csc}\left({x}\right)\:+\:\int−{sin}\left({x}\right) \\ $$$$\int{csc}\left({x}\right)\:+{cos}\left({x}\right) \\ $$$$\int{csc}\left({x}\right).\frac{{csc}\left({x}\right)+{cot}\left({x}\right)}{{csc}\left({x}\right)+{cot}\left({x}\right)}\:+\:{cos}\left({x}\right)…