Printable View

- Dec 6th 2007, 07:33 AMcazimialgebraic/trigonometric substitution

- Dec 6th 2007, 08:11 AMKrizalid
- Dec 6th 2007, 09:18 AMTwistedOne151Second one
On the second integral,

First, I think you meant

, not (you forgot to bracket the exponent).

First, we want to complete the square in the denominator:

So using , we get

From here, you can use the trigonometric substitution , which gives and

And thus

You can take it from there.

--Kevin C. - Dec 6th 2007, 09:25 AMPaulRS
Well, the third

Let thus:

Then

Now so:

Therefore: