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.
For the first one:
If $\displaystyle dw = \frac{5}{2\sqrt{x}} dx$
Then,
$\displaystyle \frac{2\sqrt{x}}{5} dw = dx$
So, we have f(x)dw = dx
but we need f(x) in terms of w.
We know that $\displaystyle w = 5\sqrt{x}$
So, $\displaystyle \sqrt{x} = \frac{w}{5}$
Using this,
$\displaystyle \frac{2(\frac{w}{5})}{5} dw = dx$
$\displaystyle \frac{2w}{25} dw = dx$
And f(w) = 2w/25
Using this in the integral:
$\displaystyle \int sin(5\sqrt{x}) dx = \int sin(w). \frac{2w}{25} dw = \int \frac{2w\ sin(w)}{25} dw$
Hence, $\displaystyle g(w) = \frac{2w\ sin(w)}{25}$
Can you continue now?