Question Number 158477 by LEKOUMA last updated on 04/Nov/21
$${I}_{{n}\:} =\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{\mathrm{2}{n}+\mathrm{1}} }{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}{dx}\:,\:{n}\geqslant\mathrm{0}\: \\ $$$${prove}\:{that}\:\forall\:{n}\geqslant\mathrm{0}\: \\ $$$$\left(\mathrm{2}{n}+\mathrm{1}\right){I}_{{n}} =\sqrt{\mathrm{2}}−\mathrm{2}{nI}_{{n}−\mathrm{1}} \\ $$
Answered by Ar Brandon last updated on 04/Nov/21
![I_n =∫_0 ^1 (x^(2n+1) /( (√(1+x^2 ))))dx, x=tanϑ⇒dx=sec^2 ϑdϑ =∫_0 ^(π/4) ((tan^(2n+1) ϑ)/(secϑ))∙sec^2 ϑdϑ=∫_0 ^(π/4) tan^(2n+1) ϑsecϑdϑ =∫_0 ^(π/4) tan^(2n) ϑ∙secϑtanϑdϑ, { ((u(ϑ)=tan^(2n) ϑ)),((v′(ϑ)=secϑtanϑ)) :} ⇒ { ((u′(ϑ)=2ntan^(2n−1) ϑsec^2 ϑ)),((v(ϑ)=secϑ)) :} =[tan^(2n) ϑsecϑ]_0 ^(π/4) −2n∫_0 ^(π/4) tan^(2n−1) sec^3 ϑdϑ =(√2)−2n∫_0 ^(π/4) tan^(2n−1) secϑdϑ−2n∫_0 ^(π/4) tan^(2n+1) ϑsecϑdϑ =(√2)−2nI_(n−1) −2nI_n ⇒ determinant ((((2n+1)I_n =(√2)−2nI_(n−1) )))](Q158490.png)
$${I}_{{n}} =\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{\mathrm{2}{n}+\mathrm{1}} }{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}{dx},\:{x}=\mathrm{tan}\vartheta\Rightarrow{dx}=\mathrm{sec}^{\mathrm{2}} \vartheta{d}\vartheta \\ $$$$\:\:\:\:\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{\mathrm{tan}^{\mathrm{2}{n}+\mathrm{1}} \vartheta}{\mathrm{sec}\vartheta}\centerdot\mathrm{sec}^{\mathrm{2}} \vartheta{d}\vartheta=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \mathrm{tan}^{\mathrm{2}{n}+\mathrm{1}} \vartheta\mathrm{sec}\vartheta{d}\vartheta \\ $$$$\:\:\:\:\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \mathrm{tan}^{\mathrm{2}{n}} \vartheta\centerdot\mathrm{sec}\vartheta\mathrm{tan}\vartheta{d}\vartheta,\:\begin{cases}{\mathrm{u}\left(\vartheta\right)=\mathrm{tan}^{\mathrm{2}{n}} \vartheta}\\{\mathrm{v}'\left(\vartheta\right)=\mathrm{sec}\vartheta\mathrm{tan}\vartheta}\end{cases} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\begin{cases}{\mathrm{u}'\left(\vartheta\right)=\mathrm{2}{n}\mathrm{tan}^{\mathrm{2}{n}−\mathrm{1}} \vartheta\mathrm{sec}^{\mathrm{2}} \vartheta}\\{\mathrm{v}\left(\vartheta\right)=\mathrm{sec}\vartheta}\end{cases} \\ $$$$\:\:\:\:\:=\left[\mathrm{tan}^{\mathrm{2}{n}} \vartheta\mathrm{sec}\vartheta\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} −\mathrm{2}{n}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \mathrm{tan}^{\mathrm{2}{n}−\mathrm{1}} \mathrm{sec}^{\mathrm{3}} \vartheta{d}\vartheta \\ $$$$\:\:\:\:\:=\sqrt{\mathrm{2}}−\mathrm{2}{n}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \mathrm{tan}^{\mathrm{2}{n}−\mathrm{1}} \mathrm{sec}\vartheta{d}\vartheta−\mathrm{2}{n}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \mathrm{tan}^{\mathrm{2}{n}+\mathrm{1}} \vartheta\mathrm{sec}\vartheta{d}\vartheta \\ $$$$\:\:\:\:\:=\sqrt{\mathrm{2}}−\mathrm{2}{nI}_{{n}−\mathrm{1}} −\mathrm{2}{nI}_{{n}} \Rightarrow\begin{array}{|c|}{\left(\mathrm{2}{n}+\mathrm{1}\right){I}_{{n}} =\sqrt{\mathrm{2}}−\mathrm{2}{nI}_{{n}−\mathrm{1}} }\\\hline\end{array} \\ $$
Answered by Ar Brandon last updated on 04/Nov/21
![I_n =∫_0 ^1 (x^(2n+1) /( (√(1+x^2 ))))dx=(1/2)∫_0 ^1 x^(2n) ∙((2x)/( (√(1+x^2 ))))dx { ((u(x)=x^(2n) )),((v′(x)=((2x)/( (√(1+x^2 )))))) :}⇒ { ((u′(x)=2nx^(2n−1) )),((v(x)=2(√(1+x^2 )))) :} I_n =[x^(2n) (√(1+x^2 ))]_0 ^1 −2n∫_0 ^1 x^(2n−1) (√(1+x^2 ))dx =(√2)−2n∫_0 ^1 ((x^(2n−1) (1+x^2 ))/( (√(1+x^2 ))))dx =(√2)−2n∫_0 ^1 (x^(2n−1) /( (√(1+x^2 ))))dx−2n∫_0 ^1 (x^(2n+1) /( (√(1+x^2 ))))dx =(√2)−2nI_(n−1) −2nI_n ⇒ determinant ((((2n+1)I_n =(√2)−2nI_(n−1) )))](Q158491.png)
$${I}_{{n}} =\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{\mathrm{2}{n}+\mathrm{1}} }{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}{dx}=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{\mathrm{2}{n}} \centerdot\frac{\mathrm{2}{x}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}{dx} \\ $$$$\begin{cases}{\mathrm{u}\left({x}\right)={x}^{\mathrm{2}{n}} }\\{\mathrm{v}'\left({x}\right)=\frac{\mathrm{2}{x}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}}\end{cases}\Rightarrow\begin{cases}{\mathrm{u}'\left({x}\right)=\mathrm{2}{nx}^{\mathrm{2}{n}−\mathrm{1}} }\\{\mathrm{v}\left({x}\right)=\mathrm{2}\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}\end{cases} \\ $$$${I}_{{n}} =\left[{x}^{\mathrm{2}{n}} \sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right]_{\mathrm{0}} ^{\mathrm{1}} −\mathrm{2}{n}\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{\mathrm{2}{n}−\mathrm{1}} \sqrt{\mathrm{1}+{x}^{\mathrm{2}} }{dx} \\ $$$$\:\:\:\:\:=\sqrt{\mathrm{2}}−\mathrm{2}{n}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{\mathrm{2}{n}−\mathrm{1}} \left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}{dx} \\ $$$$\:\:\:\:\:=\sqrt{\mathrm{2}}−\mathrm{2}{n}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{\mathrm{2}{n}−\mathrm{1}} }{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}{dx}−\mathrm{2}{n}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{\mathrm{2}{n}+\mathrm{1}} }{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}{dx} \\ $$$$\:\:\:\:\:=\sqrt{\mathrm{2}}−\mathrm{2}{nI}_{{n}−\mathrm{1}} −\mathrm{2}{nI}_{{n}} \Rightarrow\begin{array}{|c|}{\left(\mathrm{2}{n}+\mathrm{1}\right){I}_{{n}} =\sqrt{\mathrm{2}}−\mathrm{2}{nI}_{{n}−\mathrm{1}} }\\\hline\end{array} \\ $$
Commented by mnjuly1970 last updated on 05/Nov/21
$$\:{thanks}\:{alot}\:{yakhchider}... \\ $$