Question Number 83251 by mathmax by abdo last updated on 29/Feb/20
$${calculate}\:\:\int\:\:{ch}^{\mathrm{2}} \left({x}\right){sin}^{\mathrm{3}} \:{xdx} \\ $$
Commented by mathmax by abdo last updated on 29/Feb/20
$${A}\:=\int\:{ch}^{\mathrm{2}} \left({x}\right){sin}^{\mathrm{3}} \left({x}\right){dx}\:\:{we}\:{have} \\ $$$${sin}^{\mathrm{3}} {x}\:=\left(\frac{{e}^{{ix}} −{e}^{−{ix}} }{\mathrm{2}{i}}\right)^{\mathrm{3}} =−\frac{\mathrm{1}}{\mathrm{8}{i}}\left\{\:{e}^{\mathrm{3}{ix}} \:\:−\mathrm{3}{e}^{\mathrm{2}{ix}} {e}^{−{ix}} \:+\mathrm{3}\:{e}^{{ix}} {e}^{−\mathrm{2}{ix}} −{e}^{−\mathrm{3}{ix}} \right\} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{8}{i}}\left\{\:\:{e}^{\mathrm{3}{ix}} −{e}^{−\mathrm{3}{ix}} \:−\mathrm{3}\left({e}^{{ix}} −{e}^{−{ix}} \right)\right\} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{8}{i}}\left\{\:\mathrm{2}{i}\:{sin}\left(\mathrm{3}{x}\right)−\mathrm{3}\left(\mathrm{2}{i}\:{sinx}\right)\right\}\:=−\frac{\mathrm{1}}{\mathrm{4}}{sin}\left(\mathrm{3}{x}\right)+\frac{\mathrm{3}}{\mathrm{4}}{sinx}\:{also} \\ $$$${ch}^{\mathrm{2}} {x}\:=\frac{{ch}\left(\mathrm{2}{x}\right)−\mathrm{1}}{\mathrm{2}}\:=\frac{\frac{{e}^{\mathrm{2}{x}} \:+{e}^{−\mathrm{2}{x}} }{\mathrm{2}}−\mathrm{1}}{\mathrm{2}}\:=\frac{\mathrm{1}}{\mathrm{4}}{e}^{\mathrm{2}{x}} \:+\frac{\mathrm{1}}{\mathrm{4}}{e}^{−\mathrm{2}{x}} \:−\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow \\ $$$${A}\:=\frac{\mathrm{1}}{\mathrm{16}}\int\:\:\left({e}^{\mathrm{2}{x}} +{e}^{−\mathrm{2}{x}} −\mathrm{2}\right)\left(\mathrm{3}{sinx}−{sin}\left(\mathrm{3}{x}\right)\right){dx} \\ $$$$\mathrm{16}{A}\:=\int\mathrm{3}{e}^{{x}} {sinx}\:{dx}−\int\:{e}^{\mathrm{2}{x}} {sin}\left(\mathrm{3}{x}\right){dx}\:+\mathrm{3}\int\:{e}^{−\mathrm{2}{x}} {sinxdx} \\ $$$$−\int{e}^{−\mathrm{2}{x}} \:{sin}\left(\mathrm{3}{x}\right){dx}\:−\mathrm{6}\int\:{sinx}\:{dx}\:+\mathrm{2}\int\:{sin}\left(\mathrm{3}{x}\right){dx} \\ $$$$\int\:{e}^{{x}} \:{sinx}\:{dx}\:={Im}\left(\int\:{e}^{{x}} \:{e}^{{ix}} {dx}\right)\:={Im}\left(\int\:{e}^{\left(\mathrm{1}+{i}\right){x}} {dx}\right){and} \\ $$$$\int\:{e}^{\left(\mathrm{1}+{i}\right){x}} {dx}\:=\frac{\mathrm{1}}{\mathrm{1}+{i}}{e}^{\left(\mathrm{1}+{i}\right){x}} \:=\frac{\mathrm{1}−{i}}{\mathrm{2}}{e}^{{x}} \left({cosx}\:+{isinx}\right) \\ $$$$=\frac{{e}^{{x}} }{\mathrm{2}}\left({cosx}\:+{isinx}\:−{icosx}\:+{sinx}\right)\:\Rightarrow\int\:{e}^{{x}} \:{sinxdx}=\frac{{e}^{{x}} }{\mathrm{2}}\left({sinx}\:−{cosx}\right) \\ $$$$\int\:{e}^{\mathrm{2}{x}} \:{sin}\left(\mathrm{3}{x}\right){dx}\:={Im}\left(\int\:{e}^{\mathrm{2}{x}} \:{e}^{\mathrm{3}{ix}} {dx}\right)\:={Im}\left(\int\:{e}^{\left(\mathrm{2}+\mathrm{3}{i}\right){x}} {dx}\right)\: \\ $$$$\int\:\:{e}^{\left(\mathrm{2}+\mathrm{3}{i}\right){x}} {dx}\:=\frac{\mathrm{1}}{\mathrm{2}+\mathrm{3}{i}}{e}^{\left(\mathrm{2}+\mathrm{3}{i}\right){x}} \:=\frac{\mathrm{2}−\mathrm{3}{i}}{\mathrm{13}}{e}^{\mathrm{2}{x}} \left\{{cos}\left(\mathrm{3}{x}\right)+{isin}\left(\mathrm{3}{x}\right)\right\} \\ $$$$=\frac{{e}^{\mathrm{2}{x}} }{\mathrm{13}}\left\{\mathrm{2}{cos}\left(\mathrm{3}{x}\right)+\mathrm{2}{isin}\left(\mathrm{3}{x}\right)−\mathrm{3}{icos}\left(\mathrm{3}{x}\right)\:+\mathrm{3}{sin}\left(\mathrm{3}{x}\right)\right\} \\ $$$$\Rightarrow\int\:{e}^{\mathrm{2}{x}} \:{sin}\left(\mathrm{3}{x}\right){dx}\:=\frac{{e}^{\mathrm{2}{x}} }{\mathrm{13}}\left\{\:\mathrm{2}{sin}\left(\mathrm{3}{x}\right)−\mathrm{3}{cos}\left(\mathrm{3}{x}\right)\right\}…. \\ $$