How to Solve an Integration Substitution Problem?

[UrbanXrisis]

\int \frac {cos(\sqrt{x})}{\sqrt{x}}dx =?

Here's what I did:
= \int x^{-0.5}cosx^{0.5}dx
subsitute:
u= cos(\sqrt{x})
du=-sin(\sqrt{x})(0.5x^{-0.5})dx
-\frac {1}{0.5sin(\sqrt{x})}\int u du
-\frac{2}{sin(\sqrt{x})} 0.5cos^2(\sqrt{x})
-\frac{1}{sin(\sqrt{x})}cos^2(\sqrt{x})

I know I did this wrong. Any suggestions?

 

[Galileo]

\int \frac {cos(\sqrt{x})}{\sqrt{x}}dx =?
-\frac {1}{0.5sin(\sqrt{x})}\int u du

Be careful. You cannot remove \frac{1}{\sin(\sqrt(x))} from the integral, because it depends on x! (u depends on x too).

You just made a bad choice for substitution. No biggy, just try a different one. Not too many obvious option left anymore..

 

[HallsofIvy]

Well, first of all you can't just take that "sin(\sqrt{x})" out of the integral- it's a function of x!

DON'T substitute of the whole cos x^0.5, just for x^0.5.

Let u= x^0.5 so that du= 0.5x^-0.5dx and 2du= x^-0.5.

Now, its easy!

 

[pnazari]

Try substituting u=x^1/2...

oops...beat to the punch...

 

[UrbanXrisis]

is the answer 2sin(\sqrt{x})+c?

 

[pnazari]

yep...u can always check by deriving the answer and see if you get the original function.