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Question Number 142902 by greg_ed last updated on 07/Jun/21

I_n =∫_0 ^( _ (π/2)) (sin x)^n dx with integration by parts, prove that : I_(n+2) = ((n+1)/(n+2)) . I_n

$$\mathrm{I}_{{n}} =\int_{\mathrm{0}} ^{\:_{} \frac{\pi}{\mathrm{2}}} \:\left(\mathrm{sin}\:{x}\right)^{{n}} \:{dx} \\ $$$$\boldsymbol{\mathrm{with}}\:\boldsymbol{\mathrm{integration}}\:\boldsymbol{\mathrm{by}}\:\boldsymbol{\mathrm{parts}},\:\boldsymbol{\mathrm{prove}}\:\boldsymbol{\mathrm{that}}\::\: \\ $$$$\mathrm{I}_{{n}+\mathrm{2}} \:=\:\frac{{n}+\mathrm{1}}{{n}+\mathrm{2}}\:.\:\mathrm{I}_{{n}} \\ $$

Answered by qaz last updated on 07/Jun/21

I_n =∫_0 ^(π/2) sin^n xdx I_(n+2) =∫_0 ^(π/2) sin^(n+2) xdx =∫_0 ^(π/2) sin^n x(1−cos^2 x)dx =I_n −∫_0 ^(π/2) sin^n xcos xd(sin x) =I_n −(1/(n+1))∫_0 ^(π/2) cos xd(sin^(n+1) x) =I_n −(1/(n+1))(sin^(n+1) xcos x∣_0 ^(π/2) +∫_0 ^(π/2) sin^(n+2) xdx) =I_n −(I_(n+2) /(n+1)) ⇒I_(n+2) =((n+1)/(n+2))∙I_n

$$\mathrm{I}_{\mathrm{n}} =\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \mathrm{sin}\:^{\mathrm{n}} \mathrm{xdx} \\ $$$$\mathrm{I}_{\mathrm{n}+\mathrm{2}} =\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \mathrm{sin}\:^{\mathrm{n}+\mathrm{2}} \mathrm{xdx} \\ $$$$\:\:\:\:\:\:\:\:=\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \mathrm{sin}\:^{\mathrm{n}} \mathrm{x}\left(\mathrm{1}−\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}\right)\mathrm{dx} \\ $$$$\:\:\:\:\:\:\:\:=\mathrm{I}_{\mathrm{n}} −\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \mathrm{sin}\:^{\mathrm{n}} \mathrm{xcos}\:\mathrm{xd}\left(\mathrm{sin}\:\mathrm{x}\right) \\ $$$$\:\:\:\:\:\:\:\:=\mathrm{I}_{\mathrm{n}} −\frac{\mathrm{1}}{\mathrm{n}+\mathrm{1}}\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \mathrm{cos}\:\mathrm{xd}\left(\mathrm{sin}\:^{\mathrm{n}+\mathrm{1}} \mathrm{x}\right) \\ $$$$\:\:\:\:\:\:\:\:=\mathrm{I}_{\mathrm{n}} −\frac{\mathrm{1}}{\mathrm{n}+\mathrm{1}}\left(\mathrm{sin}\:^{\mathrm{n}+\mathrm{1}} \mathrm{xcos}\:\mathrm{x}\mid_{\mathrm{0}} ^{\pi/\mathrm{2}} +\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \mathrm{sin}\:^{\mathrm{n}+\mathrm{2}} \mathrm{xdx}\right) \\ $$$$\:\:\:\:\:\:\:\:=\mathrm{I}_{\mathrm{n}} −\frac{\mathrm{I}_{\mathrm{n}+\mathrm{2}} }{\mathrm{n}+\mathrm{1}} \\ $$$$\Rightarrow\mathrm{I}_{\mathrm{n}+\mathrm{2}} =\frac{\mathrm{n}+\mathrm{1}}{\mathrm{n}+\mathrm{2}}\centerdot\mathrm{I}_{\mathrm{n}} \\ $$

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