solve ∫_0 ^1 ((tan^(−1) x)/(√(1−x^2 )))dx


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Question Number 118466 by mathdave last updated on 17/Oct/20

$s o l v e$ $\int_{0}^{1} \frac{\tan^{- 1} x}{\sqrt{1 - x^{2}}} d x$
	Answered by mathmax by abdo last updated on 17/Oct/20
	$let take a try with series I =∫_0 ^1 ((arctanx)/(√(1−x^2 )))dx changement x=sint give I =∫_0 ^(π/2) ((arctan(sint))/(cost))cost dt =∫_0 ^(π/2) arctan(sint)dt =[t arctan(sint)]_0 ^(π/2) −∫_0 ^(π/2) t .((cost)/(1+sin^2 t))dt =(π^2 /8)−∫_0 ^(π/2) ((t cost)/(1+sin^2 t))dt chang. tan((t/2))=u give ∫_0 ^(π/2) ((t cost)/(1+sin^2 t))dt =∫_0 ^1 ((2 arctanu×((1−u^2 )/(1+u^2 )))/(1+((4u^2 )/(1+u^2 ))))×((2du)/(1+u^2 )) =4 ∫_0 ^1 (((1−u^2 )arctanu)/((1+u^2 )( 1+u^2 +4u^2 )))du =4 ∫_0 ^1 (((1−u^2 )arctanu)/((u^2 +1)(5u^2 +1)))du =(4/5)×((−5)/4)∫_0 ^1 (1−u^2 )arctanu{(1/(u^2 +1))−(1/(u^2 +(1/5)))}du =∫_0 ^1 (((u^2 −1)arctanu)/(1+u^2 ))du−∫_0 ^1 (((u^2 −1)arctanu)/(u^2 +(1/5)))du ∫_0 ^1 (((u^2 −1)arctanu)/(1+u^2 ))du =∫_0 ^1 (u^2 −1)arctanu(Σ_(n=0) ^∞ (−1)^n u^(2n) )du =Σ_(n=0) ^∞ (−1)^n ∫_0 ^1 (u^(2n+2) −u^(2n) )arctanu du =Σ_(n=0) ^∞ (−1)^n U_n U_n =∫_0 ^1 (u^(2n+2) −u^(2n) )arctanu du=[((u^(2n+3) /(2n+3))−(u^(2n+1) /(2n+1)))arctanu]_0 ^1 −∫_0 ^1 ((u^(2n+3) /(2n+3))−(u^(2n+1) /(2n+1)))(du/(1+u^2 )) =(π/4)((1/(2n+3))−(1/(2n+1))) +(1/(2n+1))∫_0 ^1 (u^(2n+1) /(1+u^2 ))du−(1/(2n+3))∫_0 ^1 (u^(2n+3) /(1+u^2 ))du also ∫_0 ^1 (u^(2n+1) /(1+u^2 ))du =∫_0 ^1 u^(2n+1) Σ_(p=0) ^∞ (−1)^p u^(2p) =Σ_(p=0) ^∞ (−1)^p ∫_0 ^1 u^(2n+2p+1) =Σ_(p=0) ^∞ (((−1)^p )/(2n+2p+2)) and ∫_0 ^1 (u^(2n+3) /(1+u^2 ))du =Σ_(p=0) ^∞ (((−1)^p )/(2n+2p +4)) ⇒ U_n =(π/4)((1/(2n+3))−(1/(2n+1)))+(1/(2(2n+1)))Σ_(p=0) ^∞ (((−1)^p )/(n+p+1)) −(1/(2(2n+3)))Σ_(p=0) ^∞ (((−1)^p )/(n+p+2)) ⇒ ∫_0 ^1 (((u^2 −1)arctanu)/(1+u^2 ))du =(π/4)Σ_(n=0) ^∞ (((−1)^n )/(2n+3))−(π/4)Σ_(n=0) ^∞ (((−1)^n )/(2n+1)) +(1/2)Σ_(n=0) ^∞ (Σ_(p=0) ^∞ (((−1)^p )/((2n+1)(n+p+1))))−(1/2)Σ_(n=0) ^∞ (Σ_(p=0) ^∞ (((−1)^p )/((2n+3)(n+p+2)))) ...be continued...$
	$let take a try with series I = \int_{0}^{1} \frac{arctanx}{\sqrt{1 - x^{2}}} dx changement x = sint give$ $I = \int_{0}^{\frac{π}{2}} \frac{\arctan (sint)}{cost} cost dt = \int_{0}^{\frac{π}{2}} \arctan (sint) dt$ $= {[t \arctan (sint)]}_{0}^{\frac{π}{2}} - \int_{0}^{\frac{π}{2}} t . \frac{cost}{1 + \sin^{2} t} dt$ $= \frac{π^{2}}{8} - \int_{0}^{\frac{π}{2}} \frac{t cost}{1 + \sin^{2} t} dt chang . \tan (\frac{t}{2}) = u give$ $\int_{0}^{\frac{π}{2}} \frac{t cost}{1 + \sin^{2} t} dt = \int_{0}^{1} \frac{2 arctanu \times \frac{1 - u^{2}}{1 + u^{2}}}{1 + \frac{{4 u}^{2}}{1 + u^{2}}} \times \frac{2 du}{1 + u^{2}}$ $= 4 \int_{0}^{1} \frac{(1 - u^{2}) arctanu}{(1 + u^{2}) (1 + u^{2} + {4 u}^{2})} du = 4 \int_{0}^{1} \frac{(1 - u^{2}) arctanu}{(u^{2} + 1) ({5 u}^{2} + 1)} du$ $= \frac{4}{5} \times \frac{- 5}{4} \int_{0}^{1} (1 - u^{2}) arctanu {\frac{1}{u^{2} + 1} - \frac{1}{u^{2} + \frac{1}{5}}} du$ $= \int_{0}^{1} \frac{(u^{2} - 1) arctanu}{1 + u^{2}} du - \int_{0}^{1} \frac{(u^{2} - 1) arctanu}{u^{2} + \frac{1}{5}} du$ $\int_{0}^{1} \frac{(u^{2} - 1) arctanu}{1 + u^{2}} du = \int_{0}^{1} (u^{2} - 1) arctanu (\sum_{n = 0}^{\infty} {(- 1)}^{n} u^{2 n}) du$ $= \sum_{n = 0}^{\infty} {(- 1)}^{n} \int_{0}^{1} (u^{2 n + 2} - u^{2 n}) arctanu du = \sum_{n = 0}^{\infty} {(- 1)}^{n} U_{n}$ $U_{n} = \int_{0}^{1} (u^{2 n + 2} - u^{2 n}) arctanu du = {[(\frac{u^{2 n + 3}}{2 n + 3} - \frac{u^{2 n + 1}}{2 n + 1}) arctanu]}_{0}^{1}$ $- \int_{0}^{1} (\frac{u^{2 n + 3}}{2 n + 3} - \frac{u^{2 n + 1}}{2 n + 1}) \frac{du}{1 + u^{2}}$ $= \frac{π}{4} (\frac{1}{2 n + 3} - \frac{1}{2 n + 1}) + \frac{1}{2 n + 1} \int_{0}^{1} \frac{u^{2 n + 1}}{1 + u^{2}} du - \frac{1}{2 n + 3} \int_{0}^{1} \frac{u^{2 n + 3}}{1 + u^{2}} du also$ $\int_{0}^{1} \frac{u^{2 n + 1}}{1 + u^{2}} du = \int_{0}^{1} u^{2 n + 1} \sum_{p = 0}^{\infty} {(- 1)}^{p} u^{2 p} = \sum_{p = 0}^{\infty} {(- 1)}^{p} \int_{0}^{1} u^{2 n + 2 p + 1}$ $= \sum_{p = 0}^{\infty} \frac{{(- 1)}^{p}}{2 n + 2 p + 2} and \int_{0}^{1} \frac{u^{2 n + 3}}{1 + u^{2}} du = \sum_{p = 0}^{\infty} \frac{{(- 1)}^{p}}{2 n + 2 p + 4} \Rightarrow$ $U_{n} = \frac{π}{4} (\frac{1}{2 n + 3} - \frac{1}{2 n + 1}) + \frac{1}{2 (2 n + 1)} \sum_{p = 0}^{\infty} \frac{{(- 1)}^{p}}{n + p + 1}$ $- \frac{1}{2 (2 n + 3)} \sum_{p = 0}^{\infty} \frac{{(- 1)}^{p}}{n + p + 2} \Rightarrow$ $\int_{0}^{1} \frac{(u^{2} - 1) arctanu}{1 + u^{2}} du = \frac{π}{4} \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{2 n + 3} - \frac{π}{4} \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{2 n + 1}$ $+ \frac{1}{2} \sum_{n = 0}^{\infty} (\sum_{p = 0}^{\infty} \frac{{(- 1)}^{p}}{(2 n + 1) (n + p + 1)}) - \frac{1}{2} \sum_{n = 0}^{\infty} (\sum_{p = 0}^{\infty} \frac{{(- 1)}^{p}}{(2 n + 3) (n + p + 2)})$ $. . . be continued . . .$
	Answered by TANMAY PANACEA last updated on 17/Oct/20

	$0 < \int_{0}^{1} \frac{{t a n}^{- 1} x}{\sqrt{1 - x^{2}}} d x < \int_{0}^{1} \frac{d x}{\sqrt{1 - x^{2}}}$ $0 < I <∣ {s i n}^{- 1} x ∣_{0}^{1}$ $0 < I < \frac{π}{2}$
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