find_the_fun said:
I don't understand the question:
"First make a substitution and then use integration by parts to evaluate the integral"
[math]\int sin \sqrt{x} dx[/math]
What does it have in mind by "substitution"?
$\displaystyle \begin{align*} \int{ \sin{ \left( \sqrt{x} \right) } \,\mathrm{d}x } &= \int{ \frac{2\,\sqrt{x}\,\sin{ \left( \sqrt{x} \right) } }{2\,\sqrt{x}}\,\mathrm{d}x } \end{align*}$
Now let $\displaystyle \begin{align*} u = \sqrt{x} \implies \mathrm{d}u = \frac{1}{2\,\sqrt{x}}\,\mathrm{d}x \end{align*}$ and the integral becomes
$\displaystyle \begin{align*} \int{ \frac{2\,\sqrt{x}\,\sin{ \left( \sqrt{x} \right) }}{2\,\sqrt{x}}\,\mathrm{d}x } &= \int{ 2\,u\sin{(u)}\,\mathrm{d}u} \\ &= -2\,u\cos{(u)} - \int{ -2\cos{(u)}\,\mathrm{d}u } \\ &= -2\,u\cos{(u)} + 2\int{ \cos{(u)}\,\mathrm{d}u } \\ &= -2\,u\cos{(u)} + 2\sin{(u)} + C \\ &= -2\,\sqrt{x}\,\cos{ \left( \sqrt{x} \right) } + 2\sin{ \left( \sqrt{x} \right) } + C \end{align*}$