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 byarcana last updated on 26/Jul/19

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