Integral question

1. Jun 15, 2009

transgalactic

$$\int \frac{r^2dr}{\sqrt{r^2+x^2}}=\int{rdt}\\$$
$$t=\sqrt{r^2+x^2}\\$$
$$dt=\frac{2rdr}{2\sqrt{r^2+x^2}}$$

i tried to solve it like that

2. Jun 15, 2009

Cyosis

If you do it like that the r^2 in the numerator will pose a problem. Try the substitution $r=x \sinh u$.

Last edited: Jun 15, 2009
3. Jun 15, 2009

tiny-tim

No, it would be $$\int \frac{r^2dr}{\sqrt{r^2+x^2}}=\int{\sqrt{t^2 - x^2}dt}\\$$
erm … xsinhu

4. Jun 15, 2009

Cyosis

Whoops, will fix it.

5. Jun 15, 2009

transgalactic

what do i do after
$$\int \frac{r^2dr}{\sqrt{r^2+x^2}}=\int{\sqrt{t^2 - x^2}dt}\\$$
whats the substitution
??

and i dont know hyperbolic stuff
its not on the course

6. Jun 15, 2009

Cyosis

That integral has a sign ambiguity, because if $t=\sqrt{r^2+x^2}$ then $r=\pm \sqrt{t^2-x^2}$. Which one do you take?

Do you know of a good substitution if the integrand had been $\sqrt{1-x^2}$? Try to find a trigonometric substitution for which 1-(...)^2=(...)^2.

7. Jun 15, 2009

transgalactic

there is no trigonometric substitution for it

8. Jun 15, 2009

Cyosis

There is and it is the most famous trig identity at that. Any ideas?

9. Jun 15, 2009

transgalactic

tangence goes when there is no square root on the denominator
1/(x^2+1) type

so i dont have any clue

10. Jun 15, 2009

Cyosis

Tangent works fine for the square root case as well. But you're jumping in between integrals again. It will be helpful if you stick to one integral.

11. Jun 15, 2009

Staff: Mentor

You speak with considerable authority here, but it is unwarranted. Draw a right triangle with the horizontal leg labelled x and the vertical leg labelled r and the hypotenuse labelled sqrt(r^2 + x^2). If the acute angle is labelled u, then sec u = sqrt(r^2 + x^2)/x and tan u = r/x, and sec u * du = dr/x.