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Category: Integration

dx-sin-x-1-cos-x-

Question Number 185985 by cortano1 last updated on 30/Jan/23 dxsinx(1+cosx)=? Answered by ARUNG_Brandon_MBU last updated on 30/Jan/23 I=dxsinx(1+cosx)=dx2sinx2cos3x2$$\:\:=\int\frac{\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sec}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}}{\:\sqrt{\mathrm{tan}\frac{{x}}{\mathrm{2}}}}{dx}=\int\frac{{d}\left(\mathrm{tan}\frac{{x}}{\mathrm{2}}\right)}{\:\sqrt{\mathrm{tan}\frac{{x}}{\mathrm{2}}}}=\mathrm{2}\sqrt{\mathrm{tan}\frac{{x}}{\mathrm{2}}}+{C} \

x2-2x-2-

Question Number 54896 by yusufbode1996 last updated on 14/Feb/19 (x2+2x+2) Commented by maxmathsup by imad last updated on 14/Feb/19 $${let}\:{I}\:=\int\:\sqrt{{x}^{\mathrm{2}} \:+\mathrm{2}{x}+\mathrm{2}}{dx}\:\Rightarrow{I}\:=\int\:\sqrt{\left({x}+\mathrm{1}\right)^{\mathrm{2}} \:+\mathrm{1}}{dx}\:\:{changement}\:{x}+\mathrm{1}\:={sh}\left({t}\right)\:{give} \

solve-x-Inx-dx-

Question Number 185959 by Michaelfaraday last updated on 30/Jan/23 solve:xInxdx Answered by Frix last updated on 30/Jan/23 $$\int{x}^{\mathrm{ln}\:{x}} {dx}\:\overset{{t}=\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{ln}\:{x}} {=}\:\frac{\mathrm{1}}{\:\sqrt[{\mathrm{4}}]{\mathrm{e}}}\int\mathrm{e}^{{t}^{\mathrm{2}}…