I_(n ) =∫_0 ^1 (x^(2n+1) /( (√(1+x^2 ))))dx , n≥0 prove that ∀ n≥0 (2n+1)I_n =(√2)−2nI


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Question Number 158477 by LEKOUMA last updated on 04/Nov/21

$I_{n} = \int_{0}^{1} \frac{x^{2 n + 1}}{\sqrt{1 + x^{2}}} d x, n ⩾ 0$ $p r o v e t h a t \forall n ⩾ 0$ $(2 n + 1) I_{n} = \sqrt{2} - 2 {n I}_{n - 1}$
	Answered by Ar Brandon last updated on 04/Nov/21
	$I_n =∫_0 ^1 (x^(2n+1) /( (√(1+x^2 ))))dx, x=tanϑ⇒dx=sec^2 ϑdϑ =∫_0 ^(π/4) ((tan^(2n+1) ϑ)/(secϑ))∙sec^2 ϑdϑ=∫_0 ^(π/4) tan^(2n+1) ϑsecϑdϑ =∫_0 ^(π/4) tan^(2n) ϑ∙secϑtanϑdϑ, { ((u(ϑ)=tan^(2n) ϑ)),((v′(ϑ)=secϑtanϑ)) :} ⇒ { ((u′(ϑ)=2ntan^(2n−1) ϑsec^2 ϑ)),((v(ϑ)=secϑ)) :} =[tan^(2n) ϑsecϑ]_0 ^(π/4) −2n∫_0 ^(π/4) tan^(2n−1) sec^3 ϑdϑ =(√2)−2n∫_0 ^(π/4) tan^(2n−1) secϑdϑ−2n∫_0 ^(π/4) tan^(2n+1) ϑsecϑdϑ =(√2)−2nI_(n−1) −2nI_n ⇒ determinant ((((2n+1)I_n =(√2)−2nI_(n−1) )))$
	$I_{n} = \int_{0}^{1} \frac{x^{2 n + 1}}{\sqrt{1 + x^{2}}} d x, x = \tan ϑ \Rightarrow d x = \sec^{2} ϑ d ϑ$ $= \int_{0}^{\frac{π}{4}} \frac{\tan^{2 n + 1} ϑ}{\sec ϑ} \cdot \sec^{2} ϑ d ϑ = \int_{0}^{\frac{π}{4}} \tan^{2 n + 1} ϑ \sec ϑ d ϑ$ $= \int_{0}^{\frac{π}{4}} \tan^{2 n} ϑ \cdot \sec ϑ \tan ϑ d ϑ, {\begin{cases} u (ϑ) = \tan^{2 n} ϑ \\ v^{'} (ϑ) = \sec ϑ \tan ϑ \end{cases}$ $\Rightarrow {\begin{cases} u^{'} (ϑ) = 2 n \tan^{2 n - 1} ϑ \sec^{2} ϑ \\ v (ϑ) = \sec ϑ \end{cases}$ $= {[\tan^{2 n} ϑ \sec ϑ]}_{0}^{\frac{π}{4}} - 2 n \int_{0}^{\frac{π}{4}} \tan^{2 n - 1} \sec^{3} ϑ d ϑ$ $= \sqrt{2} - 2 n \int_{0}^{\frac{π}{4}} \tan^{2 n - 1} \sec ϑ d ϑ - 2 n \int_{0}^{\frac{π}{4}} \tan^{2 n + 1} ϑ \sec ϑ d ϑ$ $= \sqrt{2} - 2 {n I}_{n - 1} - 2 {n I}_{n} \Rightarrow \begin{matrix} (2 n + 1) I_{n} = \sqrt{2} - 2 {n I}_{n - 1} \end{matrix}$
	Answered by Ar Brandon last updated on 04/Nov/21
	$I_n =∫_0 ^1 (x^(2n+1) /( (√(1+x^2 ))))dx=(1/2)∫_0 ^1 x^(2n) ∙((2x)/( (√(1+x^2 ))))dx { ((u(x)=x^(2n) )),((v′(x)=((2x)/( (√(1+x^2 )))))) :}⇒ { ((u′(x)=2nx^(2n−1) )),((v(x)=2(√(1+x^2 )))) :} I_n =[x^(2n) (√(1+x^2 ))]_0 ^1 −2n∫_0 ^1 x^(2n−1) (√(1+x^2 ))dx =(√2)−2n∫_0 ^1 ((x^(2n−1) (1+x^2 ))/( (√(1+x^2 ))))dx =(√2)−2n∫_0 ^1 (x^(2n−1) /( (√(1+x^2 ))))dx−2n∫_0 ^1 (x^(2n+1) /( (√(1+x^2 ))))dx =(√2)−2nI_(n−1) −2nI_n ⇒ determinant ((((2n+1)I_n =(√2)−2nI_(n−1) )))$
	$I_{n} = \int_{0}^{1} \frac{x^{2 n + 1}}{\sqrt{1 + x^{2}}} d x = \frac{1}{2} \int_{0}^{1} x^{2 n} \cdot \frac{2 x}{\sqrt{1 + x^{2}}} d x$ ${\begin{cases} u (x) = x^{2 n} \\ v^{'} (x) = \frac{2 x}{\sqrt{1 + x^{2}}} \end{cases} \Rightarrow {\begin{cases} u^{'} (x) = 2 {n x}^{2 n - 1} \\ v (x) = 2 \sqrt{1 + x^{2}} \end{cases}$ $I_{n} = {[x^{2 n} \sqrt{1 + x^{2}}]}_{0}^{1} - 2 n \int_{0}^{1} x^{2 n - 1} \sqrt{1 + x^{2}} d x$ $= \sqrt{2} - 2 n \int_{0}^{1} \frac{x^{2 n - 1} (1 + x^{2})}{\sqrt{1 + x^{2}}} d x$ $= \sqrt{2} - 2 n \int_{0}^{1} \frac{x^{2 n - 1}}{\sqrt{1 + x^{2}}} d x - 2 n \int_{0}^{1} \frac{x^{2 n + 1}}{\sqrt{1 + x^{2}}} d x$ $= \sqrt{2} - 2 {n I}_{n - 1} - 2 {n I}_{n} \Rightarrow \begin{matrix} (2 n + 1) I_{n} = \sqrt{2} - 2 {n I}_{n - 1} \end{matrix}$
	Commented by mnjuly1970 last updated on 05/Nov/21

	$t h a n k s a l o t y a k h c h i d e r . . .$
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