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Question Number 48178 by Abdo msup. last updated on 20/Nov/18

find ∫ ((sin(πx))/(3 +cos(2πx)))dx

$${find}\:\:\int\:\:\:\frac{{sin}\left(\pi{x}\right)}{\mathrm{3}\:+{cos}\left(\mathrm{2}\pi{x}\right)}{dx} \\ $$

Commented by Abdo msup. last updated on 25/Nov/18

A=∫ ((sin(πx))/(3+cos(2πx)))dx =_(πx =t) (1/π)∫ ((sin(t))/(3+cos(2t)))dt ∫ ((sin(t))/(3 +2cos^2 x−1)) dt =(1/π) ∫ ((sin(t))/(2+2cos^2 (x))) dx =(1/(2π)) ∫ ((sint)/(1+cos^2 t)) dt =_(cost =u) (1/(2π)) ∫ ((−du)/(1+u^2 )) =−(1/(2π)) arctan(u)=−(1/(2π)) arctan(cos(πx) +c .

$${A}=\int\:\:\:\frac{{sin}\left(\pi{x}\right)}{\mathrm{3}+{cos}\left(\mathrm{2}\pi{x}\right)}{dx}\:=_{\pi{x}\:={t}} \:\:\frac{\mathrm{1}}{\pi}\int\:\:\:\frac{{sin}\left({t}\right)}{\mathrm{3}+{cos}\left(\mathrm{2}{t}\right)}{dt} \\ $$$$\:\int\:\:\:\frac{{sin}\left({t}\right)}{\mathrm{3}\:+\mathrm{2}{cos}^{\mathrm{2}} {x}−\mathrm{1}}\:{dt}\:=\frac{\mathrm{1}}{\pi}\:\int\:\:\:\frac{{sin}\left({t}\right)}{\mathrm{2}+\mathrm{2}{cos}^{\mathrm{2}} \left({x}\right)}\:{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\pi}\:\int\:\:\:\frac{{sint}}{\mathrm{1}+{cos}^{\mathrm{2}} {t}}\:{dt}\:=_{{cost}\:={u}} \:\:\frac{\mathrm{1}}{\mathrm{2}\pi}\:\int\:\:\:\frac{−{du}}{\mathrm{1}+{u}^{\mathrm{2}} } \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}\pi}\:{arctan}\left({u}\right)=−\frac{\mathrm{1}}{\mathrm{2}\pi}\:{arctan}\left({cos}\left(\pi{x}\right)\:+{c}\:.\right. \\ $$

Answered by Smail last updated on 20/Nov/18

A=∫((sin(πx))/(3+2cos^2 (πx)−1))dx =(1/2)∫((sin(πx))/(cos^2 (πx)+1))dx t=cos(πx)⇒dt=−πsin(πx)dx A=−(1/(2π))∫(dt/(t^2 +1))=−(1/(2π))tan^(−1) (t)+C ∫((sin(πx))/(3+cos(2πx)))dx=−(1/(2π))tan^(−1) (cos(πx))+C

$${A}=\int\frac{{sin}\left(\pi{x}\right)}{\mathrm{3}+\mathrm{2}{cos}^{\mathrm{2}} \left(\pi{x}\right)−\mathrm{1}}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{sin}\left(\pi{x}\right)}{{cos}^{\mathrm{2}} \left(\pi{x}\right)+\mathrm{1}}{dx} \\ $$$${t}={cos}\left(\pi{x}\right)\Rightarrow{dt}=−\pi{sin}\left(\pi{x}\right){dx} \\ $$$${A}=−\frac{\mathrm{1}}{\mathrm{2}\pi}\int\frac{{dt}}{{t}^{\mathrm{2}} +\mathrm{1}}=−\frac{\mathrm{1}}{\mathrm{2}\pi}{tan}^{−\mathrm{1}} \left({t}\right)+{C} \\ $$$$\int\frac{{sin}\left(\pi{x}\right)}{\mathrm{3}+{cos}\left(\mathrm{2}\pi{x}\right)}{dx}=−\frac{\mathrm{1}}{\mathrm{2}\pi}{tan}^{−\mathrm{1}} \left({cos}\left(\pi{x}\right)\right)+{C} \\ $$

Commented by Abdo msup. last updated on 20/Nov/18

thanks sir.

$${thanks}\:{sir}. \\ $$

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