# Integration question

1. Jun 27, 2010

### hasan_researc

1. The problem statement, all variables and given/known data

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

2. Relevant equations

3. The attempt at a solution

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

and then??????????????????????

Last edited: Jun 27, 2010
2. Jun 27, 2010

### hasan_researc

1. The problem statement, all variables and given/known data

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

2. Relevant equations

3. 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??????????????????????
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

3. Jun 27, 2010

### 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.

4. Jun 27, 2010

### vela

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

5. Jun 27, 2010

### hasan_researc

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

6. Jun 27, 2010

### Fightfish

Nope, that is unfortunately not true at all.

7. Jun 27, 2010

### hasan_researc

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

8. Jun 27, 2010

### hasan_researc

What is an indefinite integral, then?

9. Jun 27, 2010

### zcd

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

10. Jun 27, 2010

### 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.

11. Jun 27, 2010

### Staff: Mentor

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

12. Jun 27, 2010

### Redbelly98

Staff Emeritus