Question Number 65202 by arcana last updated on 26/Jul/19 | ||
$$\int_{\mathrm{0}} ^{\pi} \:\frac{\mathrm{cos}\:\mathrm{2}\theta}{\mathrm{1}−\mathrm{2}{a}\mathrm{cos}\:\theta+{a}^{\mathrm{2}} }{d}\theta,\:{a}^{\mathrm{2}} <\mathrm{1} \\ $$$$\mathrm{answer}? \\ $$ | ||
Answered by Tanmay chaudhury last updated on 26/Jul/19 | ||
$$\boldsymbol{{formula}}\:\boldsymbol{{taken}}\:\boldsymbol{{from}}\:\boldsymbol{{S}}{l}\:{loney} \\ $$$$\frac{\mathrm{1}−{a}^{\mathrm{2}} }{\mathrm{1}−\mathrm{2}{acos}\theta+{a}^{\mathrm{2}} } \\ $$$$=\mathrm{1}+\mathrm{2}{acos}\theta+\mathrm{2}{a}^{\mathrm{2}} {cos}\mathrm{2}\theta+\mathrm{2}{a}^{\mathrm{3}} {cos}\mathrm{3}\theta+... \\ $$$${now}\: \\ $$$${I}\left({r}\right)=\int_{\mathrm{0}} ^{\pi} \frac{{cosr}\theta}{\mathrm{1}−\mathrm{2}{acos}\theta+{a}^{\mathrm{2}} }{d}\theta \\ $$$$=\frac{\mathrm{1}}{\mathrm{1}−{a}^{\mathrm{2}} }\int_{\mathrm{0}} ^{\pi} \frac{\mathrm{1}−{a}^{\mathrm{2}} }{\mathrm{1}−\mathrm{2}{acos}\theta+{a}^{\mathrm{2}} }×{cosr}\theta\:\:{d}\theta \\ $$$$=\frac{\mathrm{1}}{\mathrm{1}−{a}^{\mathrm{2}} }\int_{\mathrm{0}} ^{\pi} \left(\mathrm{1}+\mathrm{2}{acos}\theta+\mathrm{2}{a}^{\mathrm{2}} {cos}\mathrm{2}\theta+\mathrm{2}{a}^{\mathrm{3}} {cos}\mathrm{3}\theta+...\right){cosr}\theta\:{d}\theta \\ $$$$\frac{\mathrm{1}}{\mathrm{1}−{a}^{\mathrm{2}} }\int_{\mathrm{0}} ^{\pi} \left({cosr}\theta+\mathrm{2}{acos}\theta{cosr}\theta+..+\mathrm{2}{a}^{{r}} {cosr}\theta.{cosr}\theta+...\right){d}\theta \\ $$$${again}\:{formula} \\ $$$$\int_{\mathrm{0}} ^{\pi} {cosm}\theta{cosn}\theta{d}\theta=\int_{\mathrm{0}} ^{\pi} {sinm}\theta{sinn}\theta{d}\theta=\mathrm{0} \\ $$$${when}\:{m}\neq{n} \\ $$$$\boldsymbol{{if}}\:\boldsymbol{{m}}=\boldsymbol{{n}}\:\:\:\boldsymbol{{then}}\:\int_{\mathrm{0}} ^{\pi} {cosm}\theta{cosn}\theta{d}\theta=\frac{\pi}{\mathrm{2}} \\ $$$${so}\: \\ $$$${I}\left({r}\right)=\frac{\mathrm{1}}{\mathrm{1}−{a}^{\mathrm{2}} }×\left(\mathrm{0}+\mathrm{0}+...+\mathrm{2}{a}^{{r}} ×\frac{\pi}{\mathrm{2}}\right) \\ $$$${I}\left({r}\right)=\frac{{a}^{{r}} \pi}{\mathrm{1}−{a}^{\mathrm{2}} } \\ $$$${so}\:{required}\:{answer}={I}\left(\mathrm{2}\right)=\frac{{a}^{\mathrm{2}} \pi}{\mathrm{1}−{a}^{\mathrm{2}} } \\ $$$$ \\ $$ | ||
Commented by arcana last updated on 26/Jul/19 | ||
$$\mathrm{gracias}! \\ $$$$ \\ $$ | ||