Question Number 120727 by 77731 last updated on 02/Nov/20
Answered by TANMAY PANACEA last updated on 02/Nov/20
$${N}_{{r}} =\mathrm{2}+\mathrm{2}{cosx}−\mathrm{2}{cosnx}−{cos}\left({n}−\mathrm{1}\right){x}−{cos}\left({n}+\mathrm{1}\right){x} \\ $$$$=\mathrm{2}+\mathrm{2}{cosx}−\mathrm{2}{cosnx}−\mathrm{2}{cosnxcosx} \\ $$$$=\mathrm{2}\left(\mathrm{1}+{cosx}\right)−\mathrm{2}{cosnx}\left(\mathrm{1}+{cosx}\right) \\ $$$$=\mathrm{2}\left(\mathrm{1}+{cosx}\right)\left(\mathrm{1}−{cosnx}\right) \\ $$$$\frac{{N}_{{r}} }{{D}_{{r}} }=\frac{\mathrm{2}\left(\mathrm{1}+{cosx}\right)\left(\mathrm{1}−{cosnx}\right)}{\mathrm{1}−{cos}\mathrm{2}{x}}=\frac{\mathrm{2}\left(\mathrm{1}+{cosx}\right)\left(\mathrm{1}−{cosnx}\right)}{\mathrm{2}{sin}^{\mathrm{2}} {x}} \\ $$$${I}_{{n}} −{I}_{{n}−\mathrm{2}} =\int_{\mathrm{0}} ^{\pi} \left(\frac{\mathrm{1}+{cosx}}{{sin}^{\mathrm{2}} {x}}\right)\left(\left\{\mathrm{1}−{cosnx}−\mathrm{1}+{cos}\left({n}−\mathrm{2}\right){x}\right\}\:{dx}\right. \\ $$$$=\int_{\mathrm{0}} ^{\pi} \left(\frac{\mathrm{1}+{cosx}}{{sin}^{\mathrm{2}} {x}}\right)×\mathrm{2}{sin}\left({n}−\mathrm{1}\right){xsinx} \\ $$$$=\int_{\mathrm{0}} ^{\pi} \frac{\mathrm{1}+{cosx}}{{sinx}}×{sin}\left({n}−\mathrm{1}\right){xdx} \\ $$$$=\int_{\mathrm{0}} ^{\pi} \frac{\mathrm{2}{cos}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}}{\mathrm{2}{sin}\frac{{x}}{\mathrm{2}}{cos}\frac{{x}}{\mathrm{2}}}×{sin}\left({n}−\mathrm{1}\right){xdx} \\ $$$$=\int_{\mathrm{0}} ^{\pi} {cot}\frac{{x}}{\mathrm{2}}×{sin}\left({n}−\mathrm{1}\right){xdx} \\ $$$${I}_{{n}} −{I}_{{n}−\mathrm{2}} =\int_{\mathrm{0}} ^{\pi} {cot}\frac{{x}}{\mathrm{2}}{sin}\left({n}−\mathrm{1}\right){xdx} \\ $$$${wait}… \\ $$