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Math Help - help with integrals?

  1. #1
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    help with integrals?

    help with those questions??
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  2. #2
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    The first integral is,
    \int_0^{\pi} \frac{x\sin x}{1+\cos ^2 x}dx
    The function is continous on [0,\pi].
    Thus we need to find,
    \int \frac{x\sin x}{1+\cos^2 x}dx
    Express as,
    \int x\cdot \frac{\sin x}{1+\cos^2 x}dx
    Let,
    u=x und v'=\frac{\sin x}{1+\cos^2 x}.
    Thus,
    u'=1 and v=-\tan^{-1} (\cos x)
    Thus, by parts,
    -x\tan^{-1}(\cos x)+\int \tan^{-1}(\cos x)dx
    The nice thing is that though we do not know what that integral is we know that it is zero.
    Thus,
    -x\tan^{-1}(\cos x)\big|^{\pi}_0 +\int_0^{\pi}\tan^{-1}(\cos x)dx
    As mentioned that integral "dies out".
    -\pi \tan^{-1}(-1)=-\frac{3\pi^2}{4}
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  3. #3
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    The third problem.

    You have,
    \frac{x}{(x+1)^2(x^2+1)}
    The Partial Fractions Decomposition is,
    \frac{A}{x+1}+\frac{B}{x+1}+\frac{Cx+D}{x^2+1}
    With some work we find,
    \frac{1}{2} \left( \frac{1}{x^2+1} - \frac{1}{(x+1)^2} \right)

    Now you can integrate.
    The first is acrtangent.
    The second is basic substitution.
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  4. #4
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    The Last one.
    \int_0^3 x-[x]+1/3 dx

    If you think about it,
    \int_0^n [x]dx
    Where n is an integer.
    Is, 1+2+..+n-1=\frac{n(n-1)}{2}
    Because we can think of this as adding rectangles of area each with increasing area of 1. Thus it is an arithmetical sum.

    Thus,
    \int_0^3 xdx -\int_0^3 [x] dx+\int_0^3 1/3 dx
    You should be able to do these now.

    This is mine 39th Post!!!
    Last edited by ThePerfectHacker; December 31st 2006 at 05:34 PM.
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  5. #5
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    This may be cheating, but I used Excel to create a midpoint Riemann Sum with n=50.

    That gives {\Delta}x=\frac{\pi}{50}

    Some of the numbers are skewed in the display, but that doesn't matter.

    It turned out that way.


    Code:
    1	0.031416 0.000494
    2	0.094248 0.004454
    3	0.15708	0.012439
    4	0.219912	0.024571
    5	0.282743	0.041039
    6	0.345575	0.062092
    7	0.408407	0.088042
    8	0.471239	0.119259
    9	0.534071	0.156165
    10	0.596903	0.199226
    11	0.659735	0.248935
    12	0.722566	0.305786
    13	0.785398	0.37024
    14	0.84823	0.442672
    15	0.911062	0.5233
    16	0.973894	0.612103
    17	1.036726	0.70871
    18	1.099558	0.812293
    19	1.162389	0.921451
    20	1.225221	1.034128
    21	1.288053	1.147586
    22	1.350885	1.258465
    23	1.413717	1.362958
    24	1.476549	1.457091
    25	1.539381	1.537104
    26	1.602212	1.599843
    27	1.665044	1.643103
    28	1.727876	1.665837
    29	1.790708	1.668198
    30	1.85354	1.651404
    31	1.916372	1.617482
    32	1.979204	1.568957
    33	2.042035	1.508544
    34	2.104867	1.438896
    35	2.167699	1.362421
    36	2.230531	1.281182
    37	2.293363	1.196853
    38	2.356195	1.11072
    39	2.419027	1.023718
    40	2.481858	0.936468
    41	2.54469	0.849332
    42	2.607522	0.762453
    43	2.670354	0.675801
    44	2.733186	0.589206
    45	2.796018	0.502381
    46	2.85885	0.414945
    47	2.921681	0.326439
    48	2.984513	0.236332
    49	3.047345	0.144028
    50	3.110177	0.04887
    Add up the rightmost column and multiply by Pi/50 and you get

    2.467659

    Not too bad. If you run this through Maple or a TI-89 or something, you get

    2.464010027.

    Click on the links below to see the Riemann sum graphs of #1 and #2:
    Last edited by galactus; November 24th 2008 at 05:39 AM.
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  6. #6
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    Second integral can be easily solved by a substitution according to x=\frac\pi2-u.
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  7. #7
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    ...
    -x\tan^{-1}(\cos x)+\int \tan^{-1}(\cos x)dx
    The nice thing is that though we do not know what that integral is we know that it is zero.
    pardon me if this is a stupid question, but how do we know that the last integral gives zero. i can see it from the graph, but i don't suppose graphing was an option here
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  8. #8
    Grand Panjandrum
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    Quote Originally Posted by Jhevon View Post
    pardon me if this is a stupid question, but how do we know that the last integral gives zero. i can see it from the graph, but i don't suppose graphing was an option here
    It is infact a definite integral from  0 to \pi, but the integrand is antisymetric about \pi/2.

    RonL
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  9. #9
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by CaptainBlack View Post
    It is infact a definite integral from  0 to \pi, but the integrand is antisymetric about \pi/2.

    RonL
    how would you know to check something like that? is it pure experience, or is there something i should be seeing that i'm not seeing?
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  10. #10
    Grand Panjandrum
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    Quote Originally Posted by Jhevon View Post
    how would you know to check something like that? is it pure experience, or is there something i should be seeing that i'm not seeing?
    You should see that the integral is over an interval that the trig functions
    are likely to display some sort of symmetry and then look for it.

    RonL
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