## TuringTest 4 years ago Fun integral that seems to have gotten deleted, so I'm posting it again.$\int_0^\pi\frac{x\sin x}{1+\cos^2x}dx$

1. TuringTest

@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. anonymous

3. anonymous

i think i got it

4. anonymous

u integrate xsinx and sinx/cos^2x separetely

5. TuringTest

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

6. anonymous

lol nvm

7. anonymous

8. TuringTest

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

9. TuringTest

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

10. anonymous

@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. anonymous

ill post what i have

12. anonymous

$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

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

14. anonymous

ah pellet i think i see a mistake already

15. anonymous

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

16. anonymous

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

17. TuringTest

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

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. anonymous

so up until i tried to integrate by parts?

20. TuringTest

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

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

22. anonymous

ah ok, so i just evaluated that last guy wrong

23. TuringTest

yes

24. anonymous

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

25. anonymous

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

26. TuringTest

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

27. TuringTest

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

28. anonymous

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

29. TuringTest

good spirit :) see ya!