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Category: Integration

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)…

sin-2-x-cos-4x-dx-

Question Number 131068 by EDWIN88 last updated on 01/Feb/21 $$\:\int\:\mathrm{sin}\:^{\mathrm{2}} {x}\:\mathrm{cos}\:\mathrm{4}{x}\:{dx}\:=?\: \\ $$ Answered by Ar Brandon last updated on 01/Feb/21 $$\mathcal{I}=\int\mathrm{sin}^{\mathrm{2}} \mathrm{xcos4xdx}=\frac{\mathrm{1}}{\mathrm{2}}\int\left(\mathrm{1}+\mathrm{cos2x}\right)\mathrm{cos4xdx} \\ $$$$\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\int\left[\mathrm{cos4x}+\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{cos6x}+\mathrm{cos2x}\right)\right]\mathrm{dx}…

x-2-2-x-2-3-x-2-4-x-2-5-x-2-6-x-2-7-dx-

Question Number 131058 by pipin last updated on 01/Feb/21 $$\int\frac{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{2}\right)\left(\mathrm{x}^{\mathrm{2}} +\mathrm{3}\right)\left(\mathrm{x}^{\mathrm{2}} +\mathrm{4}\right)}{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{5}\right)\left(\mathrm{x}^{\mathrm{2}} +\mathrm{6}\right)\left(\mathrm{x}^{\mathrm{2}} +\mathrm{7}\right)}\:\mathrm{dx}\: \\ $$ Answered by MJS_new last updated on 01/Feb/21…