Why can't I do a simple u-substitution with $\int \sin(\sqrt{x}) dx$? When looking at this problem, I was thinking I could do a simple u-sub.
$$ \int \sin(\sqrt{x}) dx $$ 
I was thinking I could set $$ u = \sqrt(x) $$
    $$ u^2 = x $$
Then I would set $$ du = dx $$
Then I could simple integrate $\sin$ into $-\cos$
as well as $u^2$ into $\dfrac{u^3}{3} $
The I could simply sub x back into the equation
$$ -\cos\left(\frac{x^{2/3}}{3}\right) + c $$
But this yields the wrong answer. So I'm assuming I need to do integration by parts.
Edit: After reworking it with the new info, I've come out with
$$ 2\left[-\sqrt{x}\cos(\sqrt{x}) - \frac{1}{2}\left(\sin\left(\frac{x}{2}\right)\right)\right] + c $$
 A: No, not at all, we do NOT have: $u^2 = x\implies du = dx $ What we do have is $$u^2 = x \implies 2 u \,du = dx.$$ That gets you $$\int  \sin{u}\,2u\,du = 2\int u\,\sin u \,du$$
Not a straightforward substitution whose integral you can immediately evaluate, though now you can use integration by parts.

ADDED from follow-up comments:
We let $w = u \implies dw = du$. We let $dv = \sin u\, du \implies v = \int \sin u \,du = -\cos u$. Now, we express: $$2(wv - \int v \,dw)$$
$$2\int u \,\sin u\, du = -2\Big(u \cos u - \int -\cos u \,du\Big) = -2u \cos u + 2\sin u + C$$
Now, simply back substitute $u = \sqrt x$, to get: $$-2\sqrt x \cos(\sqrt x) + 2\sin (\sqrt x) + C$$

Note: your updated answer is off on the coefficient of the $\sin$ term and it's argument: that term should be $$2 \sin (\sqrt x) \quad \text{or}\quad 2(\cdots + \sin \sqrt x)$$ And we do NOT differentiate (again) the term $\sqrt x$ to get $\dfrac 1{2\sqrt x}$, and certainly not $2\left(\cdots \dfrac 12 \sin\left(\frac x2\right)\right)$. We took care of the "chain rule" component when we put $u = \sqrt x \implies u^2 = x \implies 2u\,du = dx$. If we had chosen to find $du$ by differentiating $\sqrt x$, we'd  still have gotten $2u \,du = dx$:  $$u = \sqrt x \implies du = \dfrac 1{2\sqrt x}dx \iff 2\sqrt x du = dx \iff 2 u = dx$$ which was accounted for in our early substitution.
A: Careful!  $$\frac d{dx}u^2=2u\ du$$
