shubhamsrg 3 years ago integral [ log( sqrt(1+x) + sqrt(1-x) ) dx] from x=0 to 1

1. lgbasallote

$\huge \int \limits_0^1 \log (\sqrt{1 + x} + \sqrt{1- x})$ ???

2. vikrantg4

yep i think so

3. shubhamsrg

yep.. and a dx there,,ofcorse..

4. vikrantg4

ah.. leave the typos.. concentrate on question. lol

5. hartnn

simple substitution of x=cos 2t

6. shubhamsrg

then our dx = -2sin 2t dt.. dont you think that will create a problem ? altough the sq. roots will be eliminated..

7. shubhamsrg

how'd you get that ? what i got was inside the integral (something) + log( cos t + sin t ) sin 2t dt are you missing the log there buddy ?

8. mukushla

is that $$\ln$$

9. shubhamsrg

yep..ln..

10. hartnn

i need to reconsider

11. mukushla

man this is interesting

12. mukushla

let$t=\ln(\sqrt{1+x}+\sqrt{1-x})$

13. hartnn

is there a standard formula for: $\int\limits_{}^{}\ln(1+\sqrt{1-x^2})$ if yes,then this is so simple!

14. mukushla

i think there is

15. hartnn

$=(1/2)\int\limits \limits_0^1 \log (\sqrt{1 + x} + \sqrt{1- x})^2 dx$ $=(1/2)\int\limits \limits_0^1 \log (2+2\sqrt{1-x^2})^2 dx$ $=(1/2)\int\limits \limits_0^1 \ln 2+\ln (1+\sqrt{1-x^2})^2 dx$

16. hartnn

$=(1/2)\int\limits\limits\limits \limits_0^1 \ln 2+\int\limits\limits\limits \limits_0^1\ln (1+\sqrt{1-x^2}) dx$

17. mukushla

$\ln(1+\sqrt{1-x^2})=\text{sech}^{-1}x+\ln x$this is hard to get o.O

18. hartnn
19. mukushla
20. hartnn
21. hartnn

lol,i did same exact thing :P

22. hartnn

i guess,thats integration by parts........???

23. hartnn

taking u=ln(....),v=1

24. shubhamsrg

bingo,,got it by parts! @hartnn thanks a lot!

25. hartnn

belated welcome :)