How to integrate $\int \frac{(ar)}{\sqrt{a^2 - r^2}} dr $ ?

Question

How to integrate $\int \frac{(ar)}{\sqrt{a^2 - r^2}} dr $ ?

I tried making $ u = a^2 - r^2 $ but I can't seem to get $ -a\sqrt{a^2 - r^2} $

Any help is appreciated! Thank you.

Answer 1

Hint. Make the change of variable $$ u=a^2 - r^2,\quad du=-2rdr $$ giving $$ \int \frac{(ar)}{\sqrt{a^2 - r^2}} dr=-\frac a2\int \frac{du}{\sqrt{u}} . $$

Answer 2

That is absolutely the right choice to substitute. You get

$$\int \frac{ar}{\sqrt{a^2-r^2}} dr=-\frac{a}{2} \int \frac{1}{\sqrt{u}} du$$

Answer 3

Well $u=a^2-r^2$ is a good idea. With this substitution, we get $du = -2r dr$ and finally $$\int \frac{ar}{\sqrt{a^2 - r^2}} dr= -\frac{a}{2}\int \frac{1}{\sqrt{u}} du = -a\sqrt{u} +C = -a\sqrt{a^2-r^2}+C$$ as expected.

Answer 4

Although this type of integral is usually done by trigonometric or $u$-substitution, in this case one should notice that

$$ \frac{d}{dr}\sqrt{a^2-r^2}=-\frac{r}{\sqrt{a^2-r^2}} $$

Thus one finds, without using substitution that

$$ \int\frac{ar}{\sqrt{a^2-r^2}}dr= -a\sqrt{a^2-r^2}+c$$