Question regarding integration with substition for $\int \frac{\sin({\sqrt{x}})}{\sqrt{x}}dx$ I want to understand integration by substition.
$$\int \frac{\sin\big({\sqrt{x}}\big)}{\sqrt{x}} dx$$
$u = \sqrt{x}$
$\frac{du}{dx} = \frac{1}{2\sqrt{x}}$
$du = \frac{1}{2 \sqrt{x}} dx$
$dx = 2 \sqrt{x} du$
If I insert this into the integral, we get
$$\int \frac{\sin \big(\sqrt{x} \big)}{\sqrt{x}}dx = \int \frac{\sin(u)}{u} dx = \int \frac{\sin(u)}{u} 2 \sqrt{x} du = 2 \int \sin(u) \sqrt{x} du\tag{1}$$
What am I misunderstanding?
Why do we get to
$$2 \int \sin(u) du$$
instead of
$$\int \frac{\sin(u)}{u} dx$$

 A: Note in your last step in Eq.$(1)$
$$... \int \frac{\sin(u)}{u} 2 \sqrt{x} du= 2 \int \sin(u) \sqrt{x} du\tag{1}$$
Here you need to replace $\sqrt{x}$ by $u$, so you get
$$... = \int \frac{\sin(u)}{u} ~2u~ du$$
A: You got to $2\int \sin(u)du$ as well:
$$\int \frac{\sin(\sqrt{x})}{\sqrt{x}}dx =(*)$$
$$u=\sqrt{x}$$
$$dx=2\sqrt{x}du=2udu$$
$$(*)=\int \frac{\sin(u)}{u}2udu=2\int \sin(u)du$$
A: From $u=\sqrt{x}$ we get $x
=u^2.$ Therefore $dx=2u\,du.$ Hence
$$\int {\sin\sqrt{x}\over \sqrt{x}}\,dx=\int {\sin u\over u}\, 2u\,du=2\int {\sin u}\,du$$
While substituting it is convenient not to mix variables $x$ and $u$ in one indefinite integral. Keep them separated. Similarly for $x=\varphi(u)$ we have $dx=\varphi'(u)\,du$ and
$$\int f(x)\,dx =\int f(\varphi(u))\,\varphi'(u)\,du$$
A: The issue arrises when substituting in $du=2 \sqrt x dx$.
Since $u = \sqrt x$ you can say the denominator of $u$ and $2u$ will cancel to give $2$ and so will overall give you; the integral of $2\sin(u)$.
In your working, you remove the denominator of $u$ without canceling out the $\sqrt x$ which is what causes your problem.
A: $\int \frac{\sin\big({\sqrt{x}}\big)}{\sqrt{x}} dx= -2\cos(x^\frac{1}{2})+c$ by the chain rule.
A: $$
\begin{aligned}
I & \stackrel{y=\sqrt x}{=} \int \frac{\sin y}{y}( 2 y d y)=2 \int \sin y d y=-2 \cos y+C =-2 \cos \sqrt{x}+C .
\end{aligned}
$$
