Menu Close

Category: Integration

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

Question Number 131068 by EDWIN88 last updated on 01/Feb/21 sin2xcos4xdx=? Answered by Ar Brandon last updated on 01/Feb/21 I=sin2xcos4xdx=12(1+cos2x)cos4xdx$$\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\int\left[\mathrm{cos4x}+\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{cos6x}+\mathrm{cos2x}\right)\right]\mathrm{dx}…

Question-65486

Question Number 65486 by aliesam last updated on 30/Jul/19 Answered by MJS last updated on 31/Jul/19 arctan1x2x+1dx=u=1u=x$$\:\:\:\:\:{v}=\mathrm{arctan}\:\frac{\mathrm{1}}{{x}^{\mathrm{2}} −{x}+\mathrm{1}}\:\rightarrow\:{v}'=−\frac{\mathrm{2}{x}−\mathrm{1}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)\left({x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{2}\right)}…