A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

anonymous

  • 5 years ago

what are the steps to calculate sin20 /cos70

  • This Question is Closed
  1. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    without a calculator, you mean?

  2. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yes

  3. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    gve me the steps?

  4. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Oh, lol, I thought it would be some boring addition/subtraction formulae, but I forgot a key fact: sin(x) = cos(90 - x) for all x.

  5. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Glad, too, as 20 and 70 aren't the easier to calculate directly (but it can be done without THAT much hastle)

  6. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    sin(20)=cos(90-20)=cos(70). it's pretty obvious now, isn't it?

  7. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    It was obvious before that, tbh. I could just sense it.

  8. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    LOL!! I am sure it was obvious for you.. I was talking to leonasmart!

  9. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    You're looking for a fight, aren't you Michael?

  10. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    No.. :(

  11. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    lol...thx

  12. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    so wats the answer

  13. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Michael, What about you post a real challenging problem if you have one?

  14. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Do you think you could answer it?

  15. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I posted one of my 'fun' problems yesterday, btw. No one got it, so I had to give the solution.

  16. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I used the formula that Newton wrote above to show that sin(20) is the same as cos(70), therefore, \[{\sin(20) \over \cos(70)}={\cos(70) \over \cos(70)}=1\]

  17. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    What was the problem?

  18. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ok...thx again

  19. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Oh, nothing too serious, just some simultaneous equations - is it possible to link to old topics?

  20. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    You're welcome leonasmart!

  21. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Yeah I guess.

  22. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i am going to try do one on my own now and ask u if it is correct

  23. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    btw leonas if you want proof of that sin(90-x) thing consider a RA triangle.

  24. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    OK.. We are around.

  25. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ok.....go on INewton

  26. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Congrats for the new title Newton<< (like you would care :P)

  27. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    :D I do care

  28. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    OK WHAT about sin 30/cos 60

  29. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    You do?

  30. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    the same as the old one.

  31. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    my question (but answer is there so you can cheat :( ) http://openstudy.com/updates/4dac7055d6938b0b7bb9a74d#/updates/4dac7055d6938b0b7bb9a74d hope links work here. Note, I can ask some MUCH harder ones if you want.

  32. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    OK.. I won't cheat :)

  33. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ok thx

  34. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    if I got the answer, I would post it here..K?

  35. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Ok then.

  36. 2bornot2b
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Can you please help me with my problem at http://openstudy.com/groups/mathematics#/updates/4dadd133d6938b0b6b21a94d

  37. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    about the hint you gave, what if x+y is 0?

  38. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    It makes no difference (but if they were the rest wouldn't fit anyway, so you can forget that possibility, as they cannot satisfy it). Another hint is to (if you realise why multiplying is helpful), to apply the same thing to the third one.

  39. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I already did, and I almost got the values of x and y, but the numbers don't look right to me. They are a little bit complicated. So, I am doing it again.

  40. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    No more hints Please!

  41. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    There are radicals in the answer :/ Sorry.

  42. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Oh really?

  43. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    kk

  44. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Yes. The answers aren't nice, but the idea is good (most questions of the same type (same source) are much less dirty maths though, but it is probably one of the easiest to get into so I wanted to test the water with it). It's from a non calculator exam, though, so nothing TOO nasty.

  45. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    So far, I have the two equations, which can find x and y: x+y=48/35 (1) and xy=9/35.

  46. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Yes, I'm not a big fan of irrational numbers either.

  47. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Hmm, your equations are slightly inaccurate.

  48. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    But you are meant to obtain values for (x+y) and xy first, so your method is obviously right.

  49. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I know my method is right :P .. I want my answer to be right!! just gimme a minute to check my calculations.

  50. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    hmm ok. While we're chatting, can I ask you a question: Just HOW good at Maths is Lokisan? If you can answer that.

  51. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Well I don't know that much about him.. But, from what I have seen so far, I can say he's pretty good. You can't tell if he's special since there are not so many challenging problems here.. I think you would agree with that.

  52. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I do agree with that, actually (mainly the reason I asked, as I have no way to find out). The only problem is no one on the site really wants challenging problems.

  53. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    What about we two make a change?

  54. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Haha, I may post another 'fun' question later (but this time with nicer maths and not horrid answers). But unfortunately I don't think many others care :(

  55. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    You didn't tell me what you think about lucas?

  56. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Oh, I agree, he's obviously good. I pretty much think the same about not being able to judge on here though (he 'knows' higher level maths than me (I am much younger I am almost sure) but unsure exactly how good he is).

  57. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    May I ask how old you are?

  58. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    19. Yourself?

  59. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    21

  60. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    xy=3/35.. this can't be wrong.

  61. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    It's right

  62. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    x+y=30/35?

  63. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Yep

  64. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    35xy=3 and 35x+35y=30 implies x(30-35x)=3

  65. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Am I allowed to use software to solve the quadratic equation? :(

  66. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yeah, sure (it's boring from there)

  67. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Really? Did I already get all the fun? :(

  68. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Obviously I did. :(

  69. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    :( I'll keep a lookout for a nicer one

  70. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    You know I will get two values for x and y, substitute them in two of the four equations to get a and b.

  71. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Yeah (x/y and a/b are interchangeable now, except their signs compliment each other)

  72. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I actually have a nice question I just found, but I won't have been able to type it up for ~ 10/15 minutes

  73. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    This is much nicer (it's more if you see the trick it's a very quick, nice problem, rather than grinding even after spotting how to do it)

  74. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Waiting for it :)

  75. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I wonder why nobody solved this problem. It's not that difficult.

  76. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Exactly. I guess it's longer than most though. And most people here do problems to help people more than for just doing them.

  77. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Yeah. I usually don't solve problems that take too much time.

  78. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    What major are you taking? ( assuming you already go to college )

  79. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Well University, but in England, so no 'major' - it's a slightly different system - just Maths.

  80. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    :O you don't have majors in England.. So you're going for a degree in Mathematics.

  81. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Indeed. Ugh I have the question but there is a problem in the LaTeX. Trying to find it.

  82. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    It's 10:26 PM at where I am, and I didn't get enough sleep last night. So, if it would take me too long, I might delay solving it to tomorrow morning.

  83. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Let \[I = \int^a_0 \frac{f(x)}{f(x)+f(a-x)} \mathbb{d}x\] Use a substitution to show that \[I = \int^a_0 = \frac{f(a-x)}{f(x)+f(a-x)}\mathbb{d}x \] and hence evaluate I in terms of a. Use this result to evaluate the integrals \[\int^1_0 \frac{ln(x+1)}{ln(2+x-x^2)}\mathbb{d}x\ \] and \[\int^{\frac{\pi}{2}}_0 \frac{\sin x}{\sin (x + \pi/4)} \mathbb{d}x \] Evaluate \[\int^2_{\frac{1}{2}} \frac{\sin x}{x(\sin x + \sin 1/x)} \]

  84. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    OK. I think that's right but if I see a typo I'll let you know

  85. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    There is a typo. at least one.

  86. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Where? The last integral need a dx, but apart from that..?

  87. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    2nd one

  88. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Ah, an extra = sign :/ oops

  89. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Are you sure about the second integral? I think something is wrong.

  90. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    For which part?

  91. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    f(a-x) / [(f(x) + f(a-x)] ?

  92. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    is the substitution a-x?

  93. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Yes, u = a-x (or whatever you call the variable) Note u and x are just dummy variables, so if you get the same thing in u as the one in x it is the same integral.

  94. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Oh Yeah.. That's what I was stuck at :(

  95. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    whose good i writing essays

  96. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I am, but not for you, Mary.

  97. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Hint, though Mary: It's ' Who's' (short for who is) not whose.

  98. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    can you help me write it

  99. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    plzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz

  100. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Here is the proof quickly: \[y=a-x \implies dy=-dx; I=-\int\limits_{a}^{0}{f(a-y) \over f(y)+f(a-y)}dy\] \[\implies I=\int\limits_{0}^{a}{f(a-y) \over f(y)+f(a-y)}dy\]

  101. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

  102. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Perfect. I hope you like how you use it for the next bit :D I did.

  103. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    mary try to find amistre64.. I am sure he will help you :)

  104. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    xD so true

  105. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Okay.. We will see

  106. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    My mom is calling :)

  107. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Does that mean you are going to stop :(

  108. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I'll BRB

  109. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Ah, OK.

  110. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i newton plz help me go to the english writing center plz

  111. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Back.. I will start with the first one.

  112. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    OK. you should probably consider the part 'and hence evaluate I in terms of a' (after the bit you did - it's very useful.

  113. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    And THEN do the next integrals

  114. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Hello? I really hope you saw that (and get a general solution in terms of a for the integral), otherwise I think the integrals are pretty much impossible.

  115. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    HAHA.. I should use the definition of derivatives?

  116. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Erm, no? I can give you a (sort of big) hint if you want.

  117. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    No.. Thanks.. I am just a bit sleepy. Can I leave it to tomorrow?

  118. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    OK sure. But (small hint but if you don't want one at all ignore if you can: you have two things from that are both the same integral. You have to use both of them, and you should be able to find something nicer. But cya

  119. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I'll attach a solution (image) later in this thread if you wanna check when I'm not here, but don't feel like you have to look at it or anything.

  120. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    wait!

  121. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    :( OK

  122. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    This image is the QUESTION (formatted nicely) NOT the solution.

    1 Attachment
  123. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Well, I will try it later. I will try my best not to see the solution :).. Thanks a lot anyway. It was nice talking to you. See you soon.

  124. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    OH I have a quiz tomorrow that I didn't prepare for at all :(.. Wish me luck. PEACE!!

  125. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Yeah, OK, bye. Haha, good luck.

  126. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    If you want it: See attached image

    1 Attachment
  127. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Hey I did it without looking at your answer. I think it's the right answer :)

  128. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    since \[\int\limits_{0}^{a}{f(x) \over f(x)+f(a-x)}dx=\int\limits_{0}^{a}{f(a-x) \over f(x)+f(a-x)}dx\], then \[\int\limits_{0}^{a}{f(a-x)-f(a) \over f(x)+f(a-x)}dx=\int\limits_{0}^{a}(1-2{f(x) \over f(x)+f(a-x)})dx=0\] \[\implies a-2\int\limits_{0}^{a}{f(x) \over f(x)+f(a-x)}dx=0 \] \[\implies \int\limits_{0}^{a}{f(x) \over f(x)+f(a-x)}dx=a/2\]

  129. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    To be honest, I didn't expect such a nice answer.

  130. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I will do the integrals now!!

  131. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    The first integral is very easy.. Call the integral I. \[I=\int\limits_{0}^{1}{\ln(x+1) \over \ln(2+x-x^2)}dx=\int\limits_{0}^{1}{\ln(x+1) \over \ln(x+1)+\ln(2-x)}dx\] let's check if it's in the form we're looking for: let f(x)=ln(x+1), a=1. Therefore f(a-x)=ln(1-x+1)=ln(2-x).. Hence I=a/2=0.5 :O

  132. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I almost wrote the whole solution of the 2nd one but accidentally lost it :( .. I will write it back in few minutes.

  133. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Here it is: \[\sin (x+\pi/4)=\sin x \cos(\pi/4)+\cos x (\sin \pi/4)={1 \over \sqrt2}(\sin x +\cos x)\] the integral can be written as: \[\int\limits_{0}^{\pi/2}{\sin x \over \sin x+\cos x}dx= \sqrt2\int\limits_{0}^{\pi/2} {\cos(\pi/2 -x) \over \cos (\pi/2 -x)+\cos x}dx\]

  134. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    This integral is in the form: \[\int\limits_{0}^{a}{f(a-x) \over f(x)+f(a-x)}dx\] with f(x)=cos x and a=pi/2. Hence, \[\int\limits_{0}^{\pi/2}{\sin x \over \sin(x+\pi/4)}dx=\sqrt2 {(\pi/2) \over 2}=\sqrt 2 {\pi \over 4}\]

  135. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Very nice.

  136. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    The last part is a similar idea, but you don't use the previous result, but instead the previous general idea. But yeah, very nicely done. A slight note, the value of I (a/2) may be slightly easier proved by just added the two values for I - they total to 1. So: \[2I = \int^a_01\mathbb{d}x = [x]^a_0 = a \implies I = \frac{a}{2} \]

  137. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    OH you came!! Hello :)

  138. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    LOL.. I feel SO dumb :( .. I complicated it too much

  139. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Hey (sorry was away again)

  140. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    It's fine.. I still feel SO bad about taking too long in finding that I=a/2.. Anyway, I think I figured out how to do the last integral.

  141. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Good :D But don't worry, these questions aren't meant to be easy

  142. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I used a substitution (y=1/x) and proved that: \[\int\limits_{1/a}^{a}{f(x) \over x[f(x)+f(1/x)]}dx=\int\limits_{1/a}^{a}{f(1/x) \over x[f(x)+f(1/x)]}dx\]

  143. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    :DDD

  144. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Call that integral I. So: \[2I=\int\limits_{1/a}^{a}{1 \over x}dx=\left[ \ln \left| x \right| \right]_{1/a}^{a}=\ln \left| a-1/a \right| \implies I=\ln \sqrt{a-1/a}\]

  145. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Slight error:. ln(a) and ln(1/a) are not in the same log so combine differently (easieR)

  146. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Just direct substitution with a=2 to get the value of the last integral :)

  147. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    OH :(

  148. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    You're right!! :(

  149. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    But other than that yeah

  150. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Stupid mistake I know.

  151. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I hope I won't kill myself today!!

  152. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I=ln IaI?

  153. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Yep

  154. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Thank you so much.. This Question was so amazing. I liked it and I hated myself :P

  155. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Haha, you're welcome. I'll probably find some more sometime, but have to do some other stuff first today

  156. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Good luck in whatever you have to do (if you do believe in luck) :).. I will try to find some nice problems since my weekend just started and I have plenty of time.

  157. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    has just started* right?

  158. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Yes, it is that. And actually, I just found another integration question you might like, but I have to go so won't be able to offer much help right now. See you on this thread later/tomorrow, though, probably. Or just around the site. Hope you enjoy:

    1 Attachment
  159. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Bye!

  160. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.