<s>[INST] Integral related to Pythagoras theorem ${2\over \pi}\int_{0}^{\infty}{(ab)^3\over (a^2+b^2x^2)(b^2+a^2x^2)}\mathrm dx=h^2$. Integral related to Pythagoras theorem
Triangle ABC is a right angle triangle, where Angle $ABC=90^o$.
$h$ is perpendicular to the hypotenuse AC and meet at angle ABC.
Where $a$ and $b$ are two small sides
How can I Show that h can be represented in term of this integral $(1)$

$${2\over \pi}\int_{0}^{\infty}{(ab)^3\over (a^2+b^2x^2)(b^2+a^2x^2)}\mathrm dx=h^2\tag1$$.

Any hints on this can be relate to Pythagoras theorem
Basic formulas : $AC^2=AB^2+BC^2$ and area, $A={bh\over 2}$
 [/INST] Hint: Observe if you have a right  triangle with legs $a, b$ and $h$ is as specified, then we see that
\begin{align}
\text{Area} = \frac{a\cdot b}{2} = \frac{h\cdot \sqrt{a^2+b^2}}{2} \ \ \Rightarrow \ \ h^2 = \frac{a^2b^2}{a^2+b^2}.
\end{align}
One can show that the integral
\begin{align}
\frac{2}{\pi}\int^\infty_0 \frac{(ab)^3dx}{(a^2+b^2x^2)(b^2+a^2x^2)} = \frac{(ab)^3}{ab(a^2+b^2)}= \frac{a^2b^2}{a^2+b^2} = h^2.
\end{align}
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