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    i have tried enough now and im fed up...see attachement its question 5 from STEP II June 2004..the only thing i cant work out is A and B..this is because i dont undernd the question...
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    If you make the substitution they suggest you'd get

    t + Asint + Bcost

    = t + Int[0->pi] (x + Asin x + B cos x) sin(x+t) dt

    Work out the integral on the right-hand side.

    You have functions on the LHS and RHS that are linear sums of t, sint and cost. As this is an identity, true for all t, the coefficient of sint on LHS equals that on RHS, same for t and same for cost.

    I imagine this will give you simultaneous equations in A and B which uniquely determine A and B.
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    sorry no luck would appreciate if u sould work it out
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    This question had me baffled for a long time in the exam, although once you see the trick they suggest you, which is a little cryptic it becomes easy. (whenever i use S, i mean the integral from 0 to pi btw).

    You have f(t) = t + A sin(t) + B cos(t).

    But by definiton you have f(x) = x + S f(t) sin(x+t) dt

    Hence f(x) = x + A sin(x) + B cos(x), using S f(x) sin(x) = S f(t) sin(t), likewise for cos.

    So f(x) cos(x) = x cos(x) + Asin(x)cos(x) + Bcos^2(x).

    Now integrate both sides (using A = S f(x) cos(x)), and equate coefficients, and likewise with f(x) sin(x), then solve...
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    (Original post by beauford)
    But by definiton you have f(x) = x + S f(t) sin(x+t) dt
    how did u work this out?

    can u solve as well plz
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    (Original post by integral_neo)
    how did u work this out?

    can u solve as well plz
    Well f(t) = t + S f(x) sin(x+t) dx.

    Replacing t by x and x by t, you get f(x) = x + S f(t) sin(t+x) dt
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    i cant do it!! :confused:
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    (Original post by integral_neo)
    can u solve as well plz
    Well we have S x sin(x) = pi and S x cos(x) = -2 right?

    Now using f(x) cos(x) = x cos(x) + A sin(x) cos(x) + B cos^2(x)

    = x cos(x) + A sin(2x)/2 + B(1+cos(2x))/2.

    So A = S f(x) cos(x) = S x cos(x) + A sin(2x)/2 + B(1+cos(2x))/2 = -2 + 0 + Bpi/2.

    Also f(x) sin(x) = x sin(x) + A sin^2(x) + B sin(x) cos(x)

    = x sin(x) + A(1-cos(2x))/2 + B sin(2x)/2.

    Integrating likewise gives B = pi + Api/2. Now these equations give some funny answers, not like the ones others have got, so perhaps someone can tell me where i've gone wrong (i don't think my method is the 'intended' method either, but it should still work fine)

    Actually we now have A = -2 + Bpi/2 and B = pi + Api/2 hence Bpi/2 = p^2/2 + Ap^2/4 (p = pi).

    Hence A = -2 + Bp/2 = -2 + p^2/2 + Ap^2 /4, so A(1-p^2/4) = -2 + p^2/2, hence A = -2, and so B = 0.
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    get it now atlast *feels relieved to understand the solution and then goes to sleep*

    thanx alot ur a true Trinity student
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    (Original post by integral_neo)
    get it now atlast *feels relieved to understand the solution and then goes to sleep*

    thanx alot ur a true Trinity student
    Lol i wish, that solution took me a while to get, in fact in the exam, i got 2 instead of -2 for the initial integral, and although i used the right method, i didn't get the right answer in the end, due to getting 2 earlier and some other integration mistake, i ended up with something like A = pi + pi^3-6/pi^2+4 or something daft, ah well, i did only do the q in the last 10 minutes, here's to hoping for method marks
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    Ah, I had just solved this only to find it had been done. Oh well.

    You can use the fact that the integral of Cos[x]*Sin[x] over any multiple of pi is 0 to simplify the integrals quite a lot.

    So there is no need to turn A sin(x) cos(x) into A sin(2x)/2 you should just be able to remember that and ignore them.

    But anyway, just to confirm I also get the equations A = -2 +B*pi/2 and B = 1/2(2+A)*pi, with the solution A = -2, B = 0
 
 
 
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