Intergration Problem (Substitution I think)

**jezzyjez** · January 10th 2010, 05:57 AM

I have this integral i need to do in the middle of a engineering question and I have looked in a few directions at it and cant seem to crack it!

The integral is
$\int 4r/\sqrt{2-r^2}$

First I took out factor of 4.
Then multiplyed out the square to get

$4\int(2r-r^3/2-r^2)$

And tried by parts, which didn't go to well.

So im looking at substitution making
$u = 2r -r^3$

so $du = 2-3r^2 dr$

Where do I go from here ....

**CaptainBlack** · January 10th 2010, 06:10 AM

Originally Posted by jezzyjez

I have this integral i need to do in the middle of a engineering question and I have looked in a few directions at it and cant seem to crack it!

The integral is
$\int 4r/\sqrt{2-r^2}$

First I took out factor of 4.
Then multiplyed out the square to get

$4\int(2r-r^3/2-r^2)$

And tried by parts, which didn't go to well.

So im looking at substitution making
$u = 2r -r^3$

so $du = 2-3r^2 dr$

Where do I go from here ....

What is the derivative of $-4\sqrt{2-r^2}$ ?

CB

**galactus** · January 10th 2010, 06:29 AM

Trig sub is always an old standby.

$\int\frac{4r}{\sqrt{2-r^{2}}}dr$

Let $r=\sqrt{2}sin(t), \;\ dr=\sqrt{2}cos(t)dt$

Make the subs and it whittles clean down to

$4\sqrt{2}\int sin(t)dt$

After integrating, to get it back in terms of r, make the resub:

$t=sin^{-1}(\frac{r}{\sqrt{2}})$

**rshekhar.in** · January 10th 2010, 01:49 PM

Why not substitute
$v=2-r^2$
$dr=-\frac{du}{2r}$

**galactus** · January 10th 2010, 02:13 PM

Go ahead. I was just offerening an alternative. there are numerous ways to tackle it and most other integrals

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