## anonymous 5 years ago integration

1. anonymous

$\int\limits_{0}^{.5}8dx/(4x^2+1)^2$

2. amistre64

there it went .....thought maye you fell asleep on your keyboard ;)

3. anonymous

sorry i am new took me a while to type the equation

4. amistre64

we can take out the constant "8" right?

5. anonymous

yes

6. amistre64

or do we want to use it later... if u = 4x^2 +1 ; then du = 8x dx dx = du/8x

7. amistre64

and x would equal... sqrt(u-1)/2...not sure if that helps us tho...

8. anonymous

well, this is supposed to be solved by using trig substitution. like,let 2x=tan$\theta$

9. amistre64

$\int\limits{} \frac{2}{8u^2 \sqrt{u-1}} du$

10. amistre64

we can try it by trig substitution....a little rusty, but why not :)

11. amistre64

4x^2 = tan^2(t) 2x = tan(t) ok

12. amistre64

tan^2 + 1 = sec^2 sec^2^2 = sec^4 right?

13. amistre64

1/sec^4 = cos^4 right?

14. anonymous

yes

15. amistre64

then when we pull out the 8 we get: [S] cos^4(t) dt to solve.... or did I miss somehting...

16. amistre64

it looks good to me :)

17. anonymous

wait, how did you get 4x^2 = tan^2t

18. anonymous

what i have after solving is cos^2t

19. amistre64

I simply defined it as such; since its a substitution, with the same value its good right?

20. amistre64

tan(t) = 2x -> t = tan^-1(2x)

21. anonymous

well, my teachet said a^2-x^2 : let x=asint a^2+x^2: x=atant and so on

22. amistre64

the value doesnt change, only the form it takes

23. amistre64

if we have something under a radical, then we want to substitute it with: asin(t) so that the "a" gets changed to a good value that can be taken away... there is no radical here to worry about so we dont have to worry about what "a" needs to be right?

24. amistre64

what we need here is not to get rid of a radical sign; but to transform it from one form to another.... we need tan^2 to equal 4x^2, the only way thats possible is to have tan = 2x

25. amistre64

makes sense?

26. amistre64

but I tend to miss something whenever I do this...that dt in the problem is not simply "dt" it is dt = something dx....and thats what I need to find lol

27. amistre64

tan(t) = 2x D (tan(t)) =D (2x) sec^2(t) dt = 2 dx dx = sec^2(t) dt/2 is what I should have... right?

28. anonymous

i will show u what i have. i just have problem at last

29. amistre64

so we end up with: 4 [S] cos^4 sec^2 dt

30. amistre64

ok :)

31. amistre64

4 [S] cos^2(t) dt is what it simplifies to.... im sure of it ;)

32. anonymous

yes

33. amistre64

cos^2 = 1 + sin^2..... and so on and so forth....

34. amistre64

we can reduce the power by uing the cos(2t) equations...

35. anonymous

yes what i cant do is do the final definite substitution

36. amistre64

cos(2t) = 2cos^2 - 1 1 + cos(2t) = 2cos^2 (1+cos(2t))/2 = cos^2 right?

37. amistre64

[S] 1/2 dt + [S] cos(2t)/2 dt .... looking more doable?

38. amistre64

$4 \int\limits_{} \frac{1}{2} dt + \int\limits_{} \frac{\cos(2t)}{2}dt$

39. amistre64

well that 4 times the whole lot of it :)

40. amistre64

the left side goes to 4t/2 = 2t what we need to do is concentrate in the right side...

41. amistre64

if we multiply it by 1 the value stays the same right?

42. amistre64

$\int\limits_{} \frac{2}{2} * \frac{\cos(2t)}{2} dt$

43. amistre64

lets use this to our advantage and take out the constants that dont matter.... ok?$4* \frac{1}{4} \int\limits_{} 2 \cos(2t)dt$

44. amistre64

this IS doable :)

45. amistre64

D (sin(2t)) = 2 cos(2t) right?

46. anonymous

47. amistre64

very good :) make sure you keep track of the "4" we had and use it in both these integrals right?

48. anonymous

yea. the problem with me is the final part where we substiture value. cant figure that out

49. amistre64

so lets put up our solution as we have it so far...

50. amistre64

r$F(x) = 2t + \sin(2t) + C$

51. amistre64

you agree?

52. anonymous

yes

53. amistre64

this is our key then :)

54. anonymous

wait, did we use pi/4 already?

55. amistre64

i didnt; I dont trust myself enough to change the original interval yet, so I just convert it back to values of x :)

56. anonymous

i am confused

57. amistre64

we can use this the way it is...minus the +C of course if we change the original interval to match oour substitution...which it looks like you might have down with that pi/4..... we can use that if your confident in it; but I prefer to undo the substitution of our trig with the key I provided...

58. amistre64

$2 \tan^{-1}(2x) + \frac{4x}{4x^2+1}$ from [0, .5]

59. amistre64

if I did it right, my answer is 91 :)

60. amistre64

my tan^-1(1) came back as 45 instead of pi/4..... I might have to adjust for that :)

61. amistre64

(2+pi)/2 might be more appropriate....

62. amistre64

63. amistre64

2(pi/4) + sin(2pi/4) = pi/2 + 1 = (2+pi)/2 same results as before

64. anonymous

i will put that as answer. this was a test question . n way to check it. thanks for your time

65. amistre64

I hope it was helpful ;) thanx