Question Number 48178 by Abdo msup. last updated on 20/Nov/18
$${find}\:\:\int\:\:\:\frac{{sin}\left(\pi{x}\right)}{\mathrm{3}\:+{cos}\left(\mathrm{2}\pi{x}\right)}{dx} \\ $$
Commented by Abdo msup. last updated on 25/Nov/18
$${A}=\int\:\:\:\frac{{sin}\left(\pi{x}\right)}{\mathrm{3}+{cos}\left(\mathrm{2}\pi{x}\right)}{dx}\:=_{\pi{x}\:={t}} \:\:\frac{\mathrm{1}}{\pi}\int\:\:\:\frac{{sin}\left({t}\right)}{\mathrm{3}+{cos}\left(\mathrm{2}{t}\right)}{dt} \\ $$$$\:\int\:\:\:\frac{{sin}\left({t}\right)}{\mathrm{3}\:+\mathrm{2}{cos}^{\mathrm{2}} {x}−\mathrm{1}}\:{dt}\:=\frac{\mathrm{1}}{\pi}\:\int\:\:\:\frac{{sin}\left({t}\right)}{\mathrm{2}+\mathrm{2}{cos}^{\mathrm{2}} \left({x}\right)}\:{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\pi}\:\int\:\:\:\frac{{sint}}{\mathrm{1}+{cos}^{\mathrm{2}} {t}}\:{dt}\:=_{{cost}\:={u}} \:\:\frac{\mathrm{1}}{\mathrm{2}\pi}\:\int\:\:\:\frac{−{du}}{\mathrm{1}+{u}^{\mathrm{2}} } \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}\pi}\:{arctan}\left({u}\right)=−\frac{\mathrm{1}}{\mathrm{2}\pi}\:{arctan}\left({cos}\left(\pi{x}\right)\:+{c}\:.\right. \\ $$
Answered by Smail last updated on 20/Nov/18
$${A}=\int\frac{{sin}\left(\pi{x}\right)}{\mathrm{3}+\mathrm{2}{cos}^{\mathrm{2}} \left(\pi{x}\right)−\mathrm{1}}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{sin}\left(\pi{x}\right)}{{cos}^{\mathrm{2}} \left(\pi{x}\right)+\mathrm{1}}{dx} \\ $$$${t}={cos}\left(\pi{x}\right)\Rightarrow{dt}=−\pi{sin}\left(\pi{x}\right){dx} \\ $$$${A}=−\frac{\mathrm{1}}{\mathrm{2}\pi}\int\frac{{dt}}{{t}^{\mathrm{2}} +\mathrm{1}}=−\frac{\mathrm{1}}{\mathrm{2}\pi}{tan}^{−\mathrm{1}} \left({t}\right)+{C} \\ $$$$\int\frac{{sin}\left(\pi{x}\right)}{\mathrm{3}+{cos}\left(\mathrm{2}\pi{x}\right)}{dx}=−\frac{\mathrm{1}}{\mathrm{2}\pi}{tan}^{−\mathrm{1}} \left({cos}\left(\pi{x}\right)\right)+{C} \\ $$
Commented by Abdo msup. last updated on 20/Nov/18
$${thanks}\:{sir}. \\ $$