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I-n-0-pi-2-sin-n-x-dx-Write-a-relation-between-I-n-2-and-I-n-

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Question Number 137187 by mathocean1 last updated on 30/Mar/21
I_n =∫_0 ^(π/2) sin^n x dx Write a relation between I_(n+2) and I_n .
$${I}_{{n}} =\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {sin}^{{n}} {x}\:{dx} \\ $$$${Write}\:{a}\:{relation}\:{between}\:{I}_{{n}+\mathrm{2}} \:{and}\:{I}_{{n}} . \\ $$
Answered by Dwaipayan Shikari last updated on 30/Mar/21
∫_0 ^(π/2) sin^n x dx=∫_0 ^(π/2) sin^(2(((n+1)/2))−1) (x)cos^(2((1/2))−1) (x)dx =((Γ(((n+1)/2))Γ((1/2)))/(2Γ((n/2)+1)))=J_n J_(n+2) =((Γ(((n+3)/2))Γ((1/2)))/(2Γ((n/2)+2)))=(((((n+1)/2)))/(((n/2)+1)))J_n
$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {sin}^{{n}} {x}\:{dx}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {sin}^{\mathrm{2}\left(\frac{{n}+\mathrm{1}}{\mathrm{2}}\right)−\mathrm{1}} \left({x}\right){cos}^{\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)−\mathrm{1}} \left({x}\right){dx} \\ $$$$=\frac{\Gamma\left(\frac{{n}+\mathrm{1}}{\mathrm{2}}\right)\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}{\mathrm{2}\Gamma\left(\frac{{n}}{\mathrm{2}}+\mathrm{1}\right)}={J}_{{n}} \\ $$$${J}_{{n}+\mathrm{2}} =\frac{\Gamma\left(\frac{{n}+\mathrm{3}}{\mathrm{2}}\right)\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}{\mathrm{2}\Gamma\left(\frac{{n}}{\mathrm{2}}+\mathrm{2}\right)}=\frac{\left(\frac{{n}+\mathrm{1}}{\mathrm{2}}\right)}{\left(\frac{{n}}{\mathrm{2}}+\mathrm{1}\right)}{J}_{{n}} \\ $$
Answered by mathmax by abdo last updated on 30/Mar/21
I_(n+2) =∫_0 ^(π/2) sin^n x sin^2 xdx =∫_0 ^(π/2) sin^n x(1−cos^2 x)dx =∫_0 ^(π/2) sin^n xdx −∫_0 ^(π/2) cos^2 x sin^n x dx=I_n −J J =∫_0 ^(π/2) cosx(cosx sin^n x)dx =_(by parts) [((sin^(n+1) x)/(n+1))cosx]_0 ^(π/2) −∫_0 ^(π/2) (−sinx)((sin^(n+1) x)/(n+1))dx =(1/(n+1))∫_0 ^(π/2) sin^(n+2) xdx ⇒I_(n+2 ) =I_n −(1/(n+1))I_(n+2) ⇒ (1+(1/(n+1)))I_(n+2) =I_n ⇒((n+2)/(n+1))I_(n+2) =I_n ⇒I_(n+2) =((n+1)/(n+2))I_n
$$\mathrm{I}_{\mathrm{n}+\mathrm{2}} =\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\mathrm{sin}^{\mathrm{n}} \mathrm{x}\:\mathrm{sin}^{\mathrm{2}} \:\mathrm{xdx}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\mathrm{sin}^{\mathrm{n}} \mathrm{x}\left(\mathrm{1}−\mathrm{cos}^{\mathrm{2}} \mathrm{x}\right)\mathrm{dx} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\mathrm{sin}^{\mathrm{n}} \mathrm{xdx}\:−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\mathrm{cos}^{\mathrm{2}} \mathrm{x}\:\mathrm{sin}^{\mathrm{n}} \mathrm{x}\:\mathrm{dx}=\mathrm{I}_{\mathrm{n}} −\mathrm{J} \\ $$$$\mathrm{J}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\mathrm{cosx}\left(\mathrm{cosx}\:\mathrm{sin}^{\mathrm{n}} \mathrm{x}\right)\mathrm{dx}\:=_{\mathrm{by}\:\mathrm{parts}} \:\:\left[\frac{\mathrm{sin}^{\mathrm{n}+\mathrm{1}} \mathrm{x}}{\mathrm{n}+\mathrm{1}}\mathrm{cosx}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} −\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left(−\mathrm{sinx}\right)\frac{\mathrm{sin}^{\mathrm{n}+\mathrm{1}} \mathrm{x}}{\mathrm{n}+\mathrm{1}}\mathrm{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{n}+\mathrm{1}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\mathrm{sin}^{\mathrm{n}+\mathrm{2}} \mathrm{xdx}\:\Rightarrow\mathrm{I}_{\mathrm{n}+\mathrm{2}\:} =\mathrm{I}_{\mathrm{n}} −\frac{\mathrm{1}}{\mathrm{n}+\mathrm{1}}\mathrm{I}_{\mathrm{n}+\mathrm{2}} \:\Rightarrow \\ $$$$\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{n}+\mathrm{1}}\right)\mathrm{I}_{\mathrm{n}+\mathrm{2}} =\mathrm{I}_{\mathrm{n}} \:\Rightarrow\frac{\mathrm{n}+\mathrm{2}}{\mathrm{n}+\mathrm{1}}\mathrm{I}_{\mathrm{n}+\mathrm{2}} =\mathrm{I}_{\mathrm{n}} \:\Rightarrow\mathrm{I}_{\mathrm{n}+\mathrm{2}} =\frac{\mathrm{n}+\mathrm{1}}{\mathrm{n}+\mathrm{2}}\mathrm{I}_{\mathrm{n}} \\ $$

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