Integrate: $\frac{dx}{(R^{2}+x^{2})^{3/2}}$

[hasan_researc]

Homework Statement

\int\frac{dx}{(R^{2}+x^{2})^{3/2}}

Homework Equations

The Attempt at a Solution

<br /> <br /> \mbox{Let }x = R\\tan\theta \Rightarrow dx = R\\sec^{2}d\theta\\<br /> x^{2} = R^{2}\\tan^{2}\theta\\<br /> \Rightarrow R^{2} + x^{2} = R^{2}\\( 1 + tan^{2}\theta)\\<br /> \Rightarrow R^{2} + x^{2} = R^{2}\\sec^{2}\\<br /> \Rightarrow (R^{2} + x^{2})^{3/2} = R^{3}\\sec^{3}\theta\\<br /> \\<br /> Therefore, \int\frac{dx}{(R^{2}+x^{2})^{3/2}}\\<br /> =\int\frac{R\\sec^{2}\theta}{R^{3}\\sec^{3}\theta}\\d\theta\\<br /> =\frac{1}{R^{2}}\\ \int^{\infty}_{-\infty}cos\theta\\ d\theta\\<br /> =\frac{1}{R^{2}}\stackrel{lim}{n\rightarrow\infty} \int^{N}_{-N}cos\theta\\ d\theta\\<br /> =\frac{1}{R^{2}}\stackrel{lim}{n\rightarrow\infty}(sin\theta)\mid^{N}_{-N}\\<br /> =\frac{1}{R^{2}}\stackrel{lim}{n\rightarrow\infty}(sin N + sin N)<br /> =\frac{2}{R}\stackrel{lim}{n\rightarrow\infty}sin N<br />

and then?

 

[hasan_researc]

Homework Statement

\int\frac{dx}{(R^{2}+x^{2})^{3/2}}

Homework Equations

The Attempt at a Solution

\mbox{Let }x = R\\tan\theta \Rightarrow dx = R\\sec^{2}d\theta

x^{2} = R^{2}\\tan^{2}\theta

\Rightarrow R^{2} + x^{2} = R^{2}\\( 1 + tan^{2}\theta)

\Rightarrow R^{2} + x^{2} = R^{2}\\sec^{2}\theta

\Rightarrow (R^{2} + x^{2})^{3/2} = R^{3}\\sec^{3}\theta

\mbox{Therefore, }\int\frac{dx}{(R^{2}+x^{2})^{3/2}}

=\int\frac{R\\sec^{2}\theta}{R^{3}\\sec^{3}\theta}\\d\theta

=\frac{1}{R^{2}}\\ \int^{\infty}_{-\infty}cos\theta\\ d\theta

=\frac{1}{R^{2}}\stackrel{lim}{n\rightarrow\infty} \int^{N}_{-N}cos\theta\\ d\theta

=\frac{1}{R^{2}}\stackrel{lim}{n\rightarrow\infty}(sin\theta)\mid^{N}_{-N}

=\frac{1}{R^{2}}\stackrel{lim}{n\rightarrow\infty}(sin N + sin N)

=\frac{2}{R^{2}}\stackrel{lim}{n\rightarrow\infty}sin N

and then?

 

[Char. Limit]

Why did you suddenly switch from indefinite to definite (across the entire real axis, no less) integrals partway through? I would assume that you continue to use the indefinite integral.

 

[vela]

Fixed your LaTex.

\begin{align*}<br /> \int \frac{dx}{(R^2+x^2)^{3/2}} &amp; = \int\frac{R\sec^2\theta}{R^3\sec^3\theta} d\theta \\<br /> &amp; = \frac{1}{R^2} \int^{\infty}_{-\infty}\cos\theta\,d\theta \\<br /> &amp; = \frac{1}{R^2} \lim_{N \to \infty} \int^N_{-N}\cos\theta\,d\theta \\<br /> &amp; = \frac{1}{R^2} \lim_{N \to \infty} (\sin\theta)\mid^N_{-N} \\<br /> &amp; = \frac{1}{R^2} \lim_{N \to \infty} (\sin N + \sin N) \\<br /> &amp; = \frac{2}{R^2} \lim_{N \to \infty} \sin N<br /> \end{align*}

You need to change the limits of the integral when you change variables.

 

[hasan_researc]

I thought that an indefinite integral is just a definite integral with limits that are infinity.
Isn't that true?

 

[Fightfish]

I thought that an indefinite integral is just a definite integral with limits that are infinity.
Isn't that true?

Nope, that is unfortunately not true at all.

 

[hasan_researc]

But \theta from \infty to -\infty is the same as x from \infty to -\infty. Isn't it?

 

[hasan_researc]

What is an indefinite integral, then?

 

[zcd]

\lim_{\theta\to\infty}{\tan(\theta)} doesn't exist. You have a better chance reversing the substitution before taking the limits.

 

[Fightfish]

An indefinite integral simply produces the antiderivative of the function being integrated.
\int f(x) dx = F(x) + c
where differentiating F(x) yields the function f(x).

Meanwhile, a definite integral of a function over a specified interval is equal to the difference between the values of the antiderivative at the endpoints of the interval.
\int_{a}^{b}f(x) dx = F(b) - F(a)

One can view the antiderivative as the basis with which we employ the fundamental theorem of calculus to compute definite integrals.

 

[Mark44]

I thought that an indefinite integral is just a definite integral with limits that are infinity.
Isn't that true?

As already noted, that's not true. You might be thinking of an improper integral.

 

[Redbelly98]

Moderator's note -- please continue the discussion here:

https://www.physicsforums.com/showthread.php?t=412765
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