I'm trying to solve

${\int}_{-\mathrm{\infty}}^{\mathrm{\infty}}\frac{1}{(4+{x}^{2})\sqrt{4+{x}^{2}}}dx$

Alessandra Carrillo
2022-04-08
Answered

I'm trying to solve

${\int}_{-\mathrm{\infty}}^{\mathrm{\infty}}\frac{1}{(4+{x}^{2})\sqrt{4+{x}^{2}}}dx$

You can still ask an expert for help

oanhtih6

Answered 2022-04-09
Author has **10** answers

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

$\frac{1}{\sqrt{1+{\mathrm{tan}}^{2}t}}=\left|\mathrm{cos}t\right|$

Thus the integral becomes

${\int}_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left|\mathrm{cos}t\right|,dt$

To proceed further, you need to use the following cousin of the Pythagorean theorem:

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

From this, it follows that

Thus the integral becomes

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If you integrate $y=1$, you get x. If you integrate that, you get $\frac{{x}^{2}}{2}$, following the power rule. If you continue this, integrating over and over, the antiderivatives are: 1, x, $\frac{{x}^{2}}{2}$, $\frac{{x}^{3}}{6}$, $\frac{{x}^{4}}{24}$, and so on. It follows the pattern of $\frac{{x}^{n}}{n!}$, which happens to be the infinite polynomial of ${e}^{x}$. Just by integrating $y=1$, you follow the $exp()$ function. Is there any intuitive way why this is?

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Why can't you find all antiderivatives by integrating a power series?

if $f(x)=\sum _{n=0}^{\mathrm{\infty}}\frac{{f}^{(n)}(0)}{n!}{x}^{n}$ why can't you do the following to find a general solution

$F(x)\equiv \int f(x)dx$

$F(x)=\int (\sum _{n=0}^{\mathrm{\infty}}\frac{{f}^{(n)}(0)}{n!}{x}^{n})dx=\sum _{n=0}^{\mathrm{\infty}}\frac{{f}^{(n)}(0)}{n!}(\int {x}^{n}dx)=\sum _{n=0}^{\mathrm{\infty}}\frac{{f}^{(n)}(0)}{n!}(\frac{{x}^{n+1}}{n+1})=\sum _{n=0}^{\mathrm{\infty}}\frac{{f}^{(n)}(0)}{(n+1)!}{x}^{n+1}$

I was wondering about this because I tried this approach to finding the antiderivative $\int {e}^{{x}^{2}}dx$

if $f(x)=\sum _{n=0}^{\mathrm{\infty}}\frac{{f}^{(n)}(0)}{n!}{x}^{n}$ why can't you do the following to find a general solution

$F(x)\equiv \int f(x)dx$

$F(x)=\int (\sum _{n=0}^{\mathrm{\infty}}\frac{{f}^{(n)}(0)}{n!}{x}^{n})dx=\sum _{n=0}^{\mathrm{\infty}}\frac{{f}^{(n)}(0)}{n!}(\int {x}^{n}dx)=\sum _{n=0}^{\mathrm{\infty}}\frac{{f}^{(n)}(0)}{n!}(\frac{{x}^{n+1}}{n+1})=\sum _{n=0}^{\mathrm{\infty}}\frac{{f}^{(n)}(0)}{(n+1)!}{x}^{n+1}$

I was wondering about this because I tried this approach to finding the antiderivative $\int {e}^{{x}^{2}}dx$