Question Number 71314 by mathmax by abdo last updated on 13/Oct/19

find ∫_(−1) ^1 ln((√(1+x))+(√(1−x)))dx

Commented bymathmax by abdo last updated on 13/Oct/19

let I=∫_(−1) ^1 ln((√(1+x))+(√(1−x)))dx ⇒I=2∫_0 ^1 ln((√(1+x))+(√(1−x)))dx (function even) I=_(x=cos(2t)) 2∫_(π/4) ^0 ln((√(1+cos(2t)))+(√(1−cos(2t))))(−2)sin(2t)dt =4 ∫_0 ^(π/4) ln((√2)cost +(√2)sint)sin(2t)dt =4∫_0 ^(π/4) ln((√2)((√2)cos(t−(π/4)))sin(2t)dt =4 ∫_0 ^(π/4) {ln(2)+ln(cos((π/4)−t)}sin(2t)dt =4ln(2)∫_0 ^(π/4) sin(2t)dt +4 ∫_0 ^(π/4) ln(cos((π/4)−t))sin(2t)dt =4ln(2)[−(1/2)cos(2t)]_0 ^(π/4) +4 ∫_0 ^(π/4) sin(2t)ln(cos((π/4)−t))dt =2ln(2) +4 ∫_0 ^(π/4) sin(2t)ln(cos((π/4)−t))dt changement (π/4)−t=u givd ∫_0 ^(π/4) sin(2t)ln(cos((π/4)−t))dt =−∫_0 ^(π/4) sin(2((π/4)−u))ln(cosu)(−du) =∫_0 ^(π/4) sin((π/2)−2u)ln(cosu)du =∫_0 ^(π/4) cos(2u) ln(cosu)du by parts f^′ =cos(2u) and g =ln(cosu) ⇎ ∫_0 ^(π/4) cos(2u)ln(cosu)du =[(1/2)sin(2u)ln(cosu)]_0 ^(π/4) −∫_0 ^(π/4) (1/2)sin(2u)(((−sinu)/(cosu)))du=(1/2)ln((1/(√2)))+(1/2) ∫_0 ^(π/4) ((sin(u)(2sinu cosu))/(cosu))du =−(1/2)ln((√2)) +∫_0 ^(π/4) sin^2 u du =−(1/4)ln(2)+(1/2)∫_0 ^(π/4) (1−cos(2u))du =−(1/4)ln(2)+(π/8)−(1/4)[sin(2u)]_0 ^(π/4) =−(1/4)ln(2)+(π/8)−(1/4) ⇒ I =2ln(2)−ln(2)+(π/2) −1 =ln(2)+(π/2) −1

Answered by mind is power last updated on 13/Oct/19

x=cos(2a) ⇒dx=−2sin(2a) 1+x=2cos^2 (a):1−x=2sin^2 (a) ∫_(−1) ^1 ln((√(1+x))+(√(1−x)))dx=2∫_0 ^(π/2) ln(sin(a)(√2)+cos(a)(√2))sin(2a)da ∫_0 ^(π/2) ln((√2)sin(a)+(√2)cos(a))sin(2a)da=[((−cos(2a))/2)ln((√2)sin(a)+(√2)cos(a))∫((cos(2a))/2).((cos(a)−sin(a))/(sin(a)+cos(a)))da =ln((√(2)))+(1/2)∫(cos(a)−sin(a))^2 da =ln(√2) +(1/2)∫1−sin(2a)da =ln((√2))+(1/2)((π/2))−(1/2) I=2ln((√2))+(π/2)−1

Commented byMJS last updated on 13/Oct/19

my solution is wrong (why?) but yours also the integral is approximately 1.26394

Commented bymind is power last updated on 13/Oct/19

my answer ∗2

Commented bymathmax by abdo last updated on 13/Oct/19

thanks sir.

Commented byMJS last updated on 13/Oct/19

...and I had a sign error...

Answered by MJS last updated on 13/Oct/19

I=∫ln ((√(1+x))+(√(1−x))) dx= by parts ∫u′v=uv−∫uv′ u′=1 → u=x v=ln ((√(1+x))+(√(1−x))) → v′=((1−x^2 −(√(1−x^2 )))/(2x(1−x^2 ))) =x ln ((√(1+x))+(√(1−x))) −(1/2)∫((1−x^2 −(√(1−x^2 )))/(1−x^2 ))dx −(1/2)∫((1−x^2 −(√(1−x^2 )))/(1−x^2 ))dx= [t=arcsin x → dx=(√(1−x^2 ))dt] =(1/2)∫cos t dt −(1/2)∫dt= =(1/2)sin t −(t/2)=−(x/2)+(1/2)arcsin x ⇒ I=x ln ((√(1+x))+(√(1−x))) −(x/2)+(1/2)arcsin x +C

Commented byMJS last updated on 13/Oct/19

∫_(−1) ^1 ln ((√(1+x))+(√(1−x))) dx=(π/2)−1+ln 2