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.