Solving Integrals Involving Square Roots

[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

 

[Cyosis]

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

 

[tiny-tim]

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

i tried to solve it like that

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

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

erm … xsinhu

 

[Cyosis]

Whoops, will fix it.

 

[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 don't know hyperbolic stuff
its not on the course

 

[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.

 

[transgalactic]

there is no trigonometric substitution for it

 

[Cyosis]

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

 

[transgalactic]

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

so i don't have any clue

 

[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.

 

[Mark44]

there is no trigonometric substitution for it

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.