## TuringTest Group Title Fun integral that seems to have gotten deleted, so I'm posting it again.$\int_0^\pi\frac{x\sin x}{1+\cos^2x}dx$ one year ago one year ago

1. TuringTest Group Title

@No-data this is from the cinvestav exam, and it was one I could remember how to do. @mukushla perhaps you wanna have a try?

2. jk_16 Group Title

3. mandonut Group Title

i think i got it

4. mandonut Group Title

u integrate xsinx and sinx/cos^2x separetely

5. TuringTest Group Title

not how I did it, but there more than one way to skin a cat

6. mandonut Group Title

lol nvm

7. mandonut Group Title

8. TuringTest Group Title

I'm gonna leave this up for quite a while before giving any hints :)

9. TuringTest Group Title

wow I totally didn't mean to close this question I'll repost it :S

10. eigenschmeigen Group Title

@TuringTest i might have it... although i might well have made a mistake. i have it coming out as a periodic integral after a substitution

11. eigenschmeigen Group Title

ill post what i have

12. eigenschmeigen Group Title

$u = cosx$$\frac{du}{dx} = -sinx$$dx = \frac{du}{-sinx}$ let our integral = I $I = - \int\limits^{-1}_{1} \frac{\cos^{-1}u}{1+u^2} du$$I = \int\limits^{1}_{-1} \frac{\cos^{-1}u}{1+u^2} du$$f'(u) = \frac{1}{1+u^2} , g(u) = \cos^{-1}(u)$$f(u) = \tan^{-1}u \text{ , }g'(u) = \frac{1}{\sqrt{1-u^2}}$ $\int\limits{f' \times g } \text{ } du = f \times g - \int\limits{f \times g'} \text{ } du$ $I = [\tan^{-1}(u) \times \cos^{-1}(u)]^{1}_{-1} - \int\limits{\frac{\tan^{-1}u}{\sqrt{1-u^2}}}du$$I = \frac{\pi^2}{4} - \int\limits{\frac{\tan^{-1}u}{\sqrt{1-u^2}}}du$ $\int\limits{\frac{\tan^{-1}u}{\sqrt{1-u^2}}}du = [\tan^{-1}u \times \cos^{-1}]^{1}_{-1} - \int\limits{\frac{\cos^{-1}u}{1+u^2}du}$ therefore $I = \frac{\pi^2}{4} - I$$I= \frac{\pi^2}{8}$ if this is completely wrong my excuse is tiredness!

13. TuringTest Group Title

oh you are sooooooo close! let me see if I can find your apparently minor mistake....

14. eigenschmeigen Group Title

ah pellet i think i see a mistake already

15. eigenschmeigen Group Title

last couple of lines i think i made a sign error?

16. eigenschmeigen Group Title

aaah crap i think i made a sign error, and without it i have I = I

17. TuringTest Group Title

you have it right up until $\int{\tan^{-1}u\over\sqrt{1-u^2}}du$though you did it very differently than me there is a trick to this integral, would you like a hint?

18. TuringTest Group Title

I personally think you over-complicated things by leaving them in terms of u by the way, so it took me a while to decipher your work, but it is correct :)

19. eigenschmeigen Group Title

so up until i tried to integrate by parts?

20. TuringTest Group Title

the integration by parts was correct this setup for the integral is correct, though it looks much differently than mine as you left it in terms of u and I kept mine in x

21. TuringTest Group Title

you have done everything right, but you need to integrate this last guy....

22. eigenschmeigen Group Title

ah ok, so i just evaluated that last guy wrong

23. TuringTest Group Title

yes

24. eigenschmeigen Group Title

ugh im tired, ill have one last crack at it :D

25. eigenschmeigen Group Title

also, here's one in return: $\int\limits^{1}_{0}{\frac{x}{x^3 + 1}}dx$

26. TuringTest Group Title

thanks I'll write it down, I'm leaving soon

27. TuringTest Group Title

let me know if you want a hint on the other one...

28. eigenschmeigen Group Title

no hints, but i'll have to continue it tommorow :)

29. TuringTest Group Title

good spirit :) see ya!