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    (Original post by Smaug123)
    Because the u is just a dummy variable - you'll get exactly the same result whether you integrate  I=\displaystyle\int_{-1}^{1} \dfrac{u^4\tan u -1}{1+u^2} du or  I=\displaystyle\int_{-1}^{1} \dfrac{x^4\tan x -1}{1+x^2} dx .
    original question says
    [/tex] or  I=\displaystyle\int_{-1}^{1} \dfrac{1+ x^4\tan x }

{1+x^2} dx

    Then after some substitution

     I=\displaystyle\int_{-1}^{1} \dfrac{1- x^4\tan x}{1+x^2} dx

    But because you said variable irrelevant with definite

    You can add.. left with integral of 1/x^2 +1 wich gives result of pi/2

    But if you sub in u=-x though should get
     I=\displaystyle\int_{-1}^{1} \dfrac{u^4\tan u -1}{1+u^2} du

    if you replace the u you get  I=\displaystyle\int_{-1}^{1} \dfrac{x^4\tan x -1}{1+x^2} dx

    How can you add this now- the substitution has got you nowhere( from this perspective) I know I am wrong but i just dont get this
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    (Original post by nahomyemane778)
    original question says
    [/tex] or  I=\displaystyle\int_{-1}^{1} \dfrac{1+ x^4\tan x }

{1+x^2} dx

    Then after some substitution

     I=\displaystyle\int_{-1}^{1} \dfrac{1- x^4\tan x}{1+x^2} dx

    But because you said variable irrelevant with definite

    You can add.. left with integral of 1/x^2 +1 wich gives result of pi/2

    But if you sub in u=-x though should get
     I=\displaystyle\int_{-1}^{1} \dfrac{u^4\tan u -1}{1+u^2} du

    if you replace the u you get  I=\displaystyle\int_{-1}^{1} \dfrac{x^4\tan x -1}{1+x^2} dx

    How can you add this now- the substitution has got you nowhere( from this perspective) I know I am wrong but i just dont get this
    The original I has a 1+stuff on the top, the equivalent, but rearanged I, has a 1-stuff on the top. When you add these together the stuff cancels out and you are left with 1/something.
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    (Original post by james22)
    The original I has a 1+stuff on the top, the equivalent, but rearanged I, has a 1-stuff on the top. When you add these together the stuff cancels out and you are left with 1/something.
    but when you substitute you dont get 1-stuff you get stuff-1 on top
    because you did u=-x so du/dx=-1 so you must multiply by -1 and you
    get stuff-1 not 1-stuff so when you add the numerator does not simplify
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    (Original post by nahomyemane778)
    but when you substitute you dont get 1-stuff you get stuff-1 on top
    because you did u=-x so du/dx=-1 so you must multiply by -1 and you
    get stuff-1 not 1-stuff so when you add the numerator does not simplify
    You have forgotten that during the course of the substitution, you also changed the limits of integration from {-1,1} to {1,-1}. It takes an extra minus sign to flip them back round.
    EDIT: Oh, sorry, that makes three minus signs - I confess that I misunderstood your question when you asked (I thought it was just asking why you could relabel u as x), so I didn't check the integral. It is true that the integral of \dfrac{1-x^4 \tan (x)}{x^2+1} from -1 to 1 is \dfrac{\pi}{2}, while the integral of \dfrac{u^4 \tan (u)-1}{u^2+1} from -1 to 1 is -\dfrac{\pi}{2}.
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    (Original post by Smaug123)
    You have forgotten that during the course of the substitution, you also changed the limits of integration from {-1,1} to {1,-1}. It takes an extra minus sign to flip them back round.
    Oh my goodness me. :facepalm2: I finally understand thank you! :adore:.
    What a genius trick then- how do people think of this kind of stuff?
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    (Original post by nahomyemane778)
    Oh my goodness me. :facepalm2: I finally understand thank you! :adore:.
    What a genius trick then- how do people think of this kind of stuff?
    Practice, and by noticing that almost none of the function cares about whether x is negative or positive. Also divine inspiration, I think… but it's a fairly common pattern: "make a substitution to get the negative of the original", it happens in a variety of trig contexts.
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    (Original post by james22)
    Assume, for a contradiction, that ac+bd is a prime. Using the inequality we can immediately deduce that b+d+a-c \neq 1

    Therefore b+d-a+c=1 (otherwise we have found 2 factors, contradicting the fact that ac+bd is prime).

    The equation then reduces to

    ac+bd=b+d+a-c

    which we can rearange then make use of the inequalities given to show that

    0=a(c-1)+b(d-1)+c-d) \geq 0+0+c-d=c-d>0

    which is a contradiction as required.
    Omg, sorry dude, it turns out there was a typo, no idea how I misread it over and over again! :facepalm:

    I was driving myself mad trying to find an error :lol:
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    (Original post by Jkn)
    Omg, sorry dude, it turns out there was a typo, no idea how I misread it over and over again! :facepalm:

    I was driving myself mad trying to find an error :lol:
    No problem, I thought it was a bit easy.
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    (Original post by james22)
    No problem, I thought it was a bit easy.
    Mm, what happened is that I misread it, solved it quickly (like you did) but then got really confused considering how absolutely insane this question is supposed to be. Kind of annoyed I looked at the solution though, would have been nice to see if I would have known where to start
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    Problem 264**

    Find all  x : \sin(x) = 13
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    Solution 264

     \sin(x)=\frac{\sinh (ix)}{i}=13

     ix=arsinh (13i)=\ln (13i+\sqrt{-169+1} )=ln (13+2\sqrt{42}) +ln(i)

     x=\frac{\pi}{2} +2 n \pi-i \ln (13+2\sqrt{42}) Where n is an integer.
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    Let's see who can think of some nice shortcuts.

    Problem 265**

    Is the following matrix singular?

    

\[ \left( \begin{array}{cccc}

54401 & 57668 & 15982 & 103790 \\

33223 & 26563 & 23165 & 71489 \\

36799 & 37189 & 16596 & 46152 \\

21689 & 55538 & 79922 & 51277\end{array} \right)\]
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    (Original post by FireGarden)
    Let's see who can think of some nice shortcuts.

    Problem 265**

    Is the following matrix singular?

    

\[ \left( \begin{array}{cccc}

54401 & 57668 & 15982 & 103790 \\

33223 & 26563 & 23165 & 71489 \\

36799 & 37189 & 16596 & 46152 \\

21689 & 55538 & 79922 & 51277\end{array} \right)\]
    Spoiler:
    Show

    No, it's not. Take the matrix mod 2, and find the determinant of that - it turns out to be -1, as can easily be seen by expanding around the first row and then the last row. Therefore, the determinant of the matrix is odd.
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    (Original post by FireGarden)
    Let's see who can think of some nice shortcuts.

    Problem 265**

    Is the following matrix singular?

    

\[ \left( \begin{array}{cccc}

54401 & 57668 & 15982 & 103790 \\

33223 & 26563 & 23165 & 71489 \\

36799 & 37189 & 16596 & 46152 \\

21689 & 55538 & 79922 & 51277\end{array} \right)\]
    No, as it has determinent -1559852323213094031 \neq 0
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    (Original post by james22)
    No, as it has determinent -1559852323213094031 \neq 0
    *Engineers solution
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    (Original post by bananarama2)
    *Engineers solution
    I disagree, I tihnk a sign of a true mathematician is one who can reduce a problem to one that he knows he can solve. In this case I reduced the problem to finding the determinent with the assistance of a computer. Why make things more difficult than needed?
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    Solution 265

    If you divide every row in the matrix by the leading number in the corresponding row, (1st row divided by leading number in the first row, etc) then, every number in the first column is 1 (EDIT)

    This is relatively easy to put into row echelon form. (in reduced row-echelon form, it`s the 4x4 identity matrix!)

    Take the trace of the resulting upper diagonal matrix, which is non-zero, so the matrix is non-singular.

    (OR: perform row operations until the first column is (1,0,0,0) then use co-factor expansion along the first column)
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    (Original post by Smaug123)
    Spoiler:
    Show

    No, it's not. Take the matrix mod 2, and find the determinant of that - it turns out to be -1, as can easily be seen by expanding around the first row and then the last row. Therefore, the determinant of the matrix is odd.
    The very solution i had in mind
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    Now I'm going to be a **** head and post something I think not even LotF will evaluate without a lot of head scratching. The solution I have is only using knowledge from A level.. but I doubt it'll be used. It has a nice answer, too.

    Problem 266***

    Evaluate  \displaystyle\int_{0}^{\frac{\pi  }{2}} \dfrac{1}{1+\tan(x)^{\sqrt{2}}} \ dx
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    (Original post by FireGarden)
    Now I'm going to be a **** head and post something I think not even LotF will evaluate without a lot of head scratching. The solution I have is only using knowledge from A level.. but I doubt it'll be used. It has a nice answer, too.

    Problem 266***

    Evaluate  \displaystyle\int_{0}^{\frac{\pi  }{2}} \dfrac{1}{1+\tan(x)^{\sqrt{2}}} \ dx
    This only needs A level knowledge to do it?
 
 
 
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