Using Substitution and Parts to Evaulate.

**Bracketology** · September 14th 2010, 10:22 PM

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.

**Unknown008** · September 14th 2010, 10:33 PM

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?

**Bracketology** · September 14th 2010, 10:42 PM

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

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