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Math Help - Definite integration

  1. #1
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    Definite integration

    Hi, masters.


    Can I ask a question about definite integration?

    Integration from 0 to pie (xsin(1-x^2)dx)

    I substituted (1-x^2) = t, so -1/2 integration (sint dt).
    I changed the range to from 1 to 1 - pie^2 as I used substitution.

    So, finally I got the answer -1/2(cos(1) - cos(1-pie^2)).
    But, the answer is a bit different 1/2(1 - cos(1-pie^2)).

    Can I ask what I did wrong?

    Thanks a lot.
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  2. #2
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    Re: Definite integration

    ur answer is correct -0.69056 approx value
    from books answer it is coming 0.92 which is wrong
    for verification check
    integrate xsin(1-x^2) from 0 to pi - Wolfram|Alpha
    Thanks from yanirose
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  3. #3
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    Re: Definite integration

    please.. it's pi, not pie

    $\displaystyle{\int_0^\pi}x \sin(1-x^2)~dx$

    $u=1-x^2, ~~du=-2x~dx$

    $-\dfrac 1 2\displaystyle{\int_1^{1-\pi^2}}\sin(u)~du$

    $-\dfrac 1 2 \cos(u)|^{1-\pi^2}_1$

    $-\dfrac 1 2 \left(\cos(1-\pi^2)-\cos(1)\right)$

    $\dfrac 1 2 \left(\cos(1)- \cos(1-\pi^2)\right)$

    Your mistake is that $\cos(1) \neq 1$
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  4. #4
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    Re: Definite integration

    Quote Originally Posted by romsek View Post
    please.. it's pi, not pie

    $\displaystyle{\int_0^\pi}x \sin(1-x^2)~dx$

    $u=1-x^2, ~~du=-2x~dx$

    $-\dfrac 1 2\displaystyle{\int_1^{1-\pi^2}}\sin(u)~du$

    $-\dfrac 1 2 \cos(u)|^{1-\pi^2}_1$

    $-\dfrac 1 2 \left(\cos(1-\pi^2)-\cos(1)\right)$

    $\dfrac 1 2 \left(\cos(1)- \cos(1-\pi^2)\right)$

    Your mistake is that $\cos(1) \neq 1$
    "Pie" is also a number, $\displaystyle \begin{align*} \pi \, e \approx 3.14159 \cdot 2.71828 = 8.53973 \end{align*}$ :P
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  5. #5
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    Re: Definite integration

    Sorry, but -1/2 integration sin(u)du isn't 1/1 cos(u)
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