Integration Questions

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

$$\mathrm{gracias}! \\$$$$\\$$