Question Number 101833 by bobhans last updated on 05/Jul/20

∫ _(−1)^1 (√((1+x)/(1−x))) dx ?

Answered by Dwaipayan Shikari last updated on 05/Jul/20

∫_(−1) ^1 ((1+x)/(√(1−x^2 )))dx=∫_(−1) ^1 (1/(√(1−x^2 )))+(x/(√(1−x^2 )))=[sin^(−1) x]_(−1) ^1 −(1/2)∫_(−1) ^1 ((−2x)/(√(1−x^2 )))dx  =π−(1/2)[(√(1−x^2 ))]_(−1) ^1 =π+0=π

Answered by john santu last updated on 05/Jul/20

set (√((1+x)/(1−x))) = s →dx = ((4s)/((s^2 +1)^2 )) ds  I = ∫_0 ^∞  ((4s^2 )/((s^2 +1)^2 )) ds [ by parts ]   { ((u = s)),((dv = ((4s ds)/((s^2 +1)^2 )))) :}  I= −2.∣(s/(s^2 +1)) ∣_0 ^∞ −∫−((2 ds)/(s^2 +1))  I = 0+[ 2 arctan (s) ]_0 ^∞    I = 2((π/2)) = π  (JS ⊛)

Answered by mathmax by abdo last updated on 05/Jul/20

A =∫_(−1) ^1 (√((1+x)/(1−x)))dx  changement x  =cost give   A =−∫_0 ^π  (√((1+cost)/(1−cost)))(−sint)dt =∫_0 ^π (√((2cos^2 ((t/2)))/(2sin^2 ((t/2)))))sint dt  =∫_0 ^π  ((cos((t/2)))/(sin((t/2)))) ×2sin((t/2))cos((t/2))dt =2∫_0 ^(π ) cos^2 ((t/2))dt  =∫_0 ^π (1+cost)dt =π +∫_0 ^π  cost dt =π +[sint]_0 ^π  =π+0 ⇒A =π