# Using Substitution and Parts to Evaulate.

• September 14th 2010, 11:22 PM
Bracketology
Using Substitution and Parts to Evaulate.
These problems are confusing me because the way they have each step listed out, and how they use dw. Can someone give me an explanation for these.
http://img203.imageshack.us/img203/1145/problem1v.png
http://img193.imageshack.us/img193/1518/problem2y.png
• September 14th 2010, 11:33 PM
Unknown008
For the first one:

If $dw = \frac{5}{2\sqrt{x}} dx$

Then,

$\frac{2\sqrt{x}}{5} dw = dx$

So, we have f(x)dw = dx

but we need f(x) in terms of w.

We know that $w = 5\sqrt{x}$

So, $\sqrt{x} = \frac{w}{5}$

Using this,

$\frac{2(\frac{w}{5})}{5} dw = dx$

$\frac{2w}{25} dw = dx$

And f(w) = 2w/25

Using this in the integral:

$\int sin(5\sqrt{x}) dx = \int sin(w). \frac{2w}{25} dw = \int \frac{2w\ sin(w)}{25} dw$

Hence, $g(w) = \frac{2w\ sin(w)}{25}$

Can you continue now? (Happy)
• September 14th 2010, 11:42 PM
Bracketology
Thanks! Maybe if I hadn't been staring at this problem all day I would have realized I needed to integrate back to w. I think I can do the next one now :)