Question Number 59366 by tanmay last updated on 09/May/19
$${I}_{{n}} =\int_{\mathrm{0}} ^{\pi} {sin}^{{n}} \left({x}\right){dx} \\ $$$${find}\:\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\:\:\:\frac{{I}_{{n}} }{{n}−\mathrm{1}} \\ $$
Answered by tanmay last updated on 09/May/19
$${I}_{{n}} =\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {sin}^{{n}} \left({x}\right){dx}\:\left[\int_{\mathrm{0}} ^{\mathrm{2}{a}} {f}\left({x}\right){dx}=\mathrm{2}\int_{\mathrm{0}} ^{{a}} {f}\left({x}\right)\:{when}\:{f}\left({a}−{x}\right)={f}\left({x}\right)\right. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\int_{\mathrm{0}} ^{\mathrm{2}{a}} {f}\left({x}\right){dx}=\mathrm{0}\:\:{when}\:{f}\left({a}−{x}\right)=−{f}\left({x}\right) \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left({sinx}\right)^{\mathrm{2}×\frac{{n}+\mathrm{1}}{\mathrm{2}}−\mathrm{1}} \left({cosx}\right)^{\mathrm{2}×\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}} {dx} \\ $$$$=\frac{\lceil\left(\frac{{n}+\mathrm{1}}{\mathrm{2}}\right)\lceil\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}{\lceil\left(\frac{{n}+\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\right)} \\ $$$${I}_{{n}} =\frac{\lceil\left(\frac{{n}+\mathrm{1}}{\mathrm{2}}\right)×\sqrt{\pi}}{\lceil\left(\frac{{n}}{\mathrm{2}}+\mathrm{1}\right)} \\ $$$${I}_{{n}} =\frac{\lceil\left(\frac{{n}−\mathrm{1}}{\mathrm{2}}+\mathrm{1}\right)×\sqrt{\pi}}{\lceil\left(\frac{{n}}{\mathrm{2}}+\mathrm{1}\right)} \\ $$$${I}_{{n}} =\frac{\frac{{n}−\mathrm{1}}{\mathrm{2}}}{\frac{{n}}{\mathrm{2}}}×\frac{\lceil\left(\frac{{n}−\mathrm{1}}{\mathrm{2}}\right)}{\lceil\left(\frac{{n}}{\mathrm{2}}\right)}×\sqrt{\pi}\: \\ $$$$\frac{\boldsymbol{{I}}_{{n}} }{{n}−\mathrm{1}}=\frac{\sqrt{\pi}}{{n}}×\frac{\lceil\left(\frac{{n}−\mathrm{1}}{\mathrm{2}}\right)}{\lceil\left(\frac{{n}}{\mathrm{2}}\right)} \\ $$$${formula}…\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left({sinx}\right)^{\mathrm{2}{p}−\mathrm{1}} \left({cosx}\right)^{\mathrm{2}{q}−\mathrm{1}} {dx} \\ $$$$\:\:\:=\frac{\left.\lceil\left({p}\right)\lceil{q}\right)}{\lceil\left({p}+{q}\right)}\:{and}\:\lceil\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\sqrt{\pi}\: \\ $$$${pls}\:{wait}\:…{Tanmay} \\ $$
Commented by tanmay last updated on 09/May/19
$${somd}\:{one}\:{ask}\:{me}\:{to}\:{solve}\:{it}\:{so}\:{i}\:{posted}\:{it}… \\ $$