1. ## integration by substitution

I'm completely lost with this.

Need to calculate the integral $\displaystyle \int {\frac{{\sin (\sqrt u )}}{{\sqrt u }}du}$ using the substitution $\displaystyle {\rm x = }\sqrt {\rm u}$

As far as I am getting is this
$\displaystyle \begin{array}{l} = \int {\frac{{\sin (x)}}{x}du} \\ = \int {\frac{1}{x}\sin (x)du} \\ \end{array}$

and that $\displaystyle dx = \frac{1}{2}u^{\frac{{ - 1}}{2}} du$

Could someone solve and try and explain method. thanks

2. Just sub $\displaystyle z = \sqrt{u}$ You should get ∫2sin(z)

3. Sorry but I still do not understand.

4. $\displaystyle \int \frac{\sin u^{\frac{1}{2}}}{u^{\frac{1}{2}}} \ du$

$\displaystyle {\color{blue}x} ={\color{blue} u^{\frac{1}{2}}} \ \Rightarrow \ dx = \frac{1}{2}u^{-\frac{1}{2}} \ du \iff {\color{red}2 dx} = {\color{red}\frac{1}{u^{\frac{1}{2}}} \ du}$

So: $\displaystyle \int \frac{\sin {\color{blue}u^{\frac{1}{2}}}}{{\color{red}u^{\fra c{1}{2}}}} \ {\color{red} du} = \int \sin {\color{blue}x} ({\color{red}2 \ dx}) = 2\int \sin x \ dx$