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Is there a way to further simplify or solve this integral?

$$\int_0^L{\sqrt{1+\left(\frac{k}{P(x)}\right)^2}dx}$$ where $k$, $L$ are constant; $P(x)$ is a function defined over $0$ to $L$.

$P(x)$ takes the form:

$$P(x)=\frac{(y_2-y_1)*2}{L}x+y_1\quad \textrm{when }x\in\left[0-\frac{L}{2}\right]$$ $$P(x)=\frac{(y_1-y_2)*2}{L}\left(x-\frac{L}{2}\right)+y_2\quad \textrm{when }x\in\left[\frac{L}{2}-L\right]$$ where $y_1$ and $y_2$ are also constants.

I have tried out MATHEMATICA, but it does not give me a useful result.

Rebellos
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Zhang Ze
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  • What does $x = [ \frac{L}{2} - L ]$ mean ? Also, an integral is not an equation. – Rebellos Dec 11 '18 at 19:38
  • $x=[\frac{L}{2}-L]$ means x takes value from $\frac{L}{2}$ to $L$.

    And yes, I will edit the description. I want to have a simplified form.

     – Zhang Ze Dec 11 '18 at 20:08
  • @Rebellos perhaps, $x\in\left[\frac{L}{2},L\right]$? The hyphen seems ambiguous and is not standard interval notation. – Dando18 Dec 11 '18 at 20:19

1 Answers1

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For reducing the noise let $\alpha=\frac{2(y_2-y_1)}{L}$, then we split up the integral from $[0,\frac{L}{2}]$ and $[\frac{L}{2},L]$, after a substitution we get; $$I=\int_0^{\frac{L}{2}}\sqrt{1+(\frac{k}{\alpha x +y_1})^2}+\int_{\frac{L}{2}}^{L}\sqrt{1+(\frac{k}{-\alpha (x-\frac{L}{2}) +y_2})^2}=$$ $$\int_0^{\frac{L}{2}}\sqrt{1+(\frac{k}{\alpha x+y_1})^2}dx+\int_0^{\frac{L}{2}}\sqrt{1+(\frac{k}{-\alpha x+y_2})^2}dx$$ Letting $$\frac{\alpha x+y_1}{k} = u \qquad \frac{-\alpha x+y_1}{k} = v$$ $$dx=\frac{k}{\alpha}du \qquad dx=\frac{k}{-\alpha}dv$$ $$I=\frac{2k}{y_2-y_1}\int_{\frac{y_1}{k}}^{\frac{y_2}{k}}\sqrt{\frac{x^2+1}{x^2}}dx$$ Try mathematica now.

Andrew Kovács
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