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Question Number 59366 by tanmay last updated on 09/May/19
I_n =∫_0 ^π sin^n (x)dx find Σ_(n=2) ^∞ (I_n /(n−1))
$${I}_{{n}} =\int_{\mathrm{0}} ^{\pi} {sin}^{{n}} \left({x}\right){dx} \\ $$$${find}\:\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\:\:\:\frac{{I}_{{n}} }{{n}−\mathrm{1}} \\ $$
Answered by tanmay last updated on 09/May/19
I_n =2∫_0 ^(π/2) sin^n (x)dx [∫_0 ^(2a) f(x)dx=2∫_0 ^a f(x) when f(a−x)=f(x) ∫_0 ^(2a) f(x)dx=0 when f(a−x)=−f(x) =2∫_0 ^(π/2) (sinx)^(2×((n+1)/2)−1) (cosx)^(2×(1/2)−1) dx =((⌈(((n+1)/2))⌈((1/2)))/(⌈(((n+1)/2)+(1/2)))) I_n =((⌈(((n+1)/2))×(√π))/(⌈((n/2)+1))) I_n =((⌈(((n−1)/2)+1)×(√π))/(⌈((n/2)+1))) I_n =(((n−1)/2)/(n/2))×((⌈(((n−1)/2)))/(⌈((n/2))))×(√π) (I_n /(n−1))=((√π)/n)×((⌈(((n−1)/2)))/(⌈((n/2)))) formula...2∫_0 ^(π/2) (sinx)^(2p−1) (cosx)^(2q−1) dx =((⌈(p)⌈q))/(⌈(p+q))) and ⌈((1/2))=(√π) pls wait ...Tanmay
$${I}_{{n}} =\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {sin}^{{n}} \left({x}\right){dx}\:\left[\int_{\mathrm{0}} ^{\mathrm{2}{a}} {f}\left({x}\right){dx}=\mathrm{2}\int_{\mathrm{0}} ^{{a}} {f}\left({x}\right)\:{when}\:{f}\left({a}−{x}\right)={f}\left({x}\right)\right. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\int_{\mathrm{0}} ^{\mathrm{2}{a}} {f}\left({x}\right){dx}=\mathrm{0}\:\:{when}\:{f}\left({a}−{x}\right)=−{f}\left({x}\right) \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left({sinx}\right)^{\mathrm{2}×\frac{{n}+\mathrm{1}}{\mathrm{2}}−\mathrm{1}} \left({cosx}\right)^{\mathrm{2}×\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}} {dx} \\ $$$$=\frac{\lceil\left(\frac{{n}+\mathrm{1}}{\mathrm{2}}\right)\lceil\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}{\lceil\left(\frac{{n}+\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\right)} \\ $$$${I}_{{n}} =\frac{\lceil\left(\frac{{n}+\mathrm{1}}{\mathrm{2}}\right)×\sqrt{\pi}}{\lceil\left(\frac{{n}}{\mathrm{2}}+\mathrm{1}\right)} \\ $$$${I}_{{n}} =\frac{\lceil\left(\frac{{n}−\mathrm{1}}{\mathrm{2}}+\mathrm{1}\right)×\sqrt{\pi}}{\lceil\left(\frac{{n}}{\mathrm{2}}+\mathrm{1}\right)} \\ $$$${I}_{{n}} =\frac{\frac{{n}−\mathrm{1}}{\mathrm{2}}}{\frac{{n}}{\mathrm{2}}}×\frac{\lceil\left(\frac{{n}−\mathrm{1}}{\mathrm{2}}\right)}{\lceil\left(\frac{{n}}{\mathrm{2}}\right)}×\sqrt{\pi}\: \\ $$$$\frac{\boldsymbol{{I}}_{{n}} }{{n}−\mathrm{1}}=\frac{\sqrt{\pi}}{{n}}×\frac{\lceil\left(\frac{{n}−\mathrm{1}}{\mathrm{2}}\right)}{\lceil\left(\frac{{n}}{\mathrm{2}}\right)} \\ $$$${formula}…\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left({sinx}\right)^{\mathrm{2}{p}−\mathrm{1}} \left({cosx}\right)^{\mathrm{2}{q}−\mathrm{1}} {dx} \\ $$$$\:\:\:=\frac{\left.\lceil\left({p}\right)\lceil{q}\right)}{\lceil\left({p}+{q}\right)}\:{and}\:\lceil\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\sqrt{\pi}\: \\ $$$${pls}\:{wait}\:…{Tanmay} \\ $$
Commented by tanmay last updated on 09/May/19
somd one ask me to solve it so i posted it...
$${somd}\:{one}\:{ask}\:{me}\:{to}\:{solve}\:{it}\:{so}\:{i}\:{posted}\:{it}… \\ $$

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