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I-n-0-1-x-2n-1-1-x-2-dx-n-0-prove-that-n-0-2n-1-I-n-2-2nI-n-1-




Question Number 158477 by LEKOUMA last updated on 04/Nov/21
I_(n ) =∫_0 ^1 (x^(2n+1) /( (√(1+x^2 ))))dx , n≥0   prove that ∀ n≥0   (2n+1)I_n =(√2)−2nI_(n−1)
In=01x2n+11+x2dx,n0provethatn0(2n+1)In=22nIn1
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) )))
In=01x2n+11+x2dx,x=tanϑdx=sec2ϑdϑ=0π4tan2n+1ϑsecϑsec2ϑdϑ=0π4tan2n+1ϑsecϑdϑ=0π4tan2nϑsecϑtanϑdϑ,{u(ϑ)=tan2nϑv(ϑ)=secϑtanϑ{u(ϑ)=2ntan2n1ϑsec2ϑv(ϑ)=secϑ=[tan2nϑsecϑ]0π42n0π4tan2n1sec3ϑdϑ=22n0π4tan2n1secϑdϑ2n0π4tan2n+1ϑsecϑdϑ=22nIn12nIn(2n+1)In=22nIn1
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) )))
In=01x2n+11+x2dx=1201x2n2x1+x2dx{u(x)=x2nv(x)=2x1+x2{u(x)=2nx2n1v(x)=21+x2In=[x2n1+x2]012n01x2n11+x2dx=22n01x2n1(1+x2)1+x2dx=22n01x2n11+x2dx2n01x2n+11+x2dx=22nIn12nIn(2n+1)In=22nIn1
Commented by mnjuly1970 last updated on 05/Nov/21
 thanks alot yakhchider...
thanksalotyakhchider

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