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Math Help - intergration

  1. #1
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    intergration

    noting that conclude that
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by rmpatel5 View Post
    noting that conclude that
    f(x)=\frac{1+x^2}{1+x^4}. So we can see that f(-x)=f(x) so f(x) is even. Thus

    \begin{aligned}\int_0^1\frac{1+x^2}{1+x^4}dx&=\fra  c{1}{2}\cdot2\int_0^1\frac{1+x^2}{1+x^4}dx\\<br />
&=\frac{1}{2}\int_{-1}^{1}\frac{1+x^2}{1+x^4}dx\end{aligned}

    So using the given info we may rewrite this as

    \begin{aligned}\frac{1}{2}\int_{-1}^{1}\frac{1+x^2}{1+x^4}dx&=\frac{1}{4}\int_{-1}^1\frac{2(1+x^2)}{1+x^4}dx\\<br />
&=\frac{1}{4}\int_{-1}^1\bigg[\frac{1}{1+\sqrt{2}x+x^2}+\frac{1}{1-\sqrt{2}x+x^2}\bigg]<br />
\end{aligned}

    Now note that \frac{1}{1+\sqrt{2}(-x)+x^2}=\frac{1}{1-\sqrt{2}x+x^2}

    So they are refelctions of each other over the y-axis, so while they may not be equal, they have the same area underneath.

    so

    \begin{aligned}\frac{1}{4}\int_{-1}^1\bigg[\frac{1}{1+\sqrt{2}x+x^2}+\frac{1}{1-\sqrt{2}x+x^2}\bigg]dx&=\frac{1}{4}\int_{-1}^{1}\frac{2}{1+\sqrt{2}x+x^2}dx\\<br />
&=\frac{1}{2}\int_{-1}^{1}\frac{dx}{1+\sqrt{2}x+x^2}dx\\<br />
&=\frac{1}{2}\int_{-1}^1\frac{dx}{1-\sqrt{2}x+x^2}dx\\<br />
&=\int_0^1\frac{1+x^2}{1+x^4}dx<br />
\end{aligned}
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