# I'm trying to solve $$\displaystyle{\int_{{-\infty}}^{{\infty}}}{\frac{{{1}}}{{{\left({4}+{x}^{{2}}\right)}\sqrt{{{4}+{x}^{{2}}}}}}}{\left.{d}{x}\right.}$$

Alessandra Carrillo 2022-04-08 Answered
I'm trying to solve
${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{1}{\left(4+{x}^{2}\right)\sqrt{4+{x}^{2}}}dx$
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oanhtih6
Therefore, after substitution, the integral becomes
${\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\frac{1}{\sqrt{1+{\mathrm{tan}}^{2}t}},dt$
To proceed further, you need to use the following cousin of the Pythagorean theorem:
$1+{\mathrm{tan}}^{2}t={\mathrm{sec}}^{2}t$
This identity is extremely important and useful in practice -- not the least in manipulating integrals like this. One should be reminded of this identity whenever one comes across an expression like $\sqrt{1+{\mathrm{tan}}^{2}t}$ or $\sqrt{1+{x}^{2}}$. By the way, the proof of this identity is based on the standard Pythagorean theorem:
$1+{\mathrm{tan}}^{2}t=1+\frac{{\mathrm{sin}}^{2}t}{{\mathrm{cos}}^{2}t}=\frac{{\mathrm{cos}}^{2}t+{\mathrm{sin}}^{2}t}{{\mathrm{cos}}^{2}t}=\frac{1}{{\mathrm{cos}}^{2}t}={\mathrm{sec}}^{2}t$
From this, it follows that
$\frac{1}{\sqrt{1+{\mathrm{tan}}^{2}t}}=|\mathrm{cos}t|$
Thus the integral becomes
${\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}|\mathrm{cos}t|,dt$