A couple of integrals...here's where I am with the first

[cheeze]

The problem is:

?(sin?x)/?x dx

I let u=?x
Then du=1/2x^(-1/2)dx
Then 2du=x^(-1/2)dx
Then -2du=x^(1/2)dx=?xdx

My TI-89 is telling me that the answer is -2 cos?x

So the integral I'm at is:
?(sinu)/u -2du

Which I don't believe will get me to what the TI-89 is telling me.

I know I am messed up somewhere.... Please help

 

[galactus]

\(\displaystyle \int\frac{sin(\sqrt{x})}{\sqrt{x}}dx\)

Let \(\displaystyle u=\sqrt{x}, \;\ 2du=\frac{1}{\sqrt{x}}dx\)

\(\displaystyle 2\int sin(u)du\)

There ya' go. Easy now?. You were on the right track with the subs.

I see where you made your mistake.

Here:

Then 2du=x^(-1/2)dx
Then -2du=x^(1/2)dx=?xdx