Need help with integral and change the variable I have

$$\displaystyle \int_0^\pi \sqrt{x} ~\cos x ~dx$$

and I need to make change of the variable $u = \sin x$.
$\int_0^\pi\sqrt{x}\cos x~dx=\int_0^\pi x^{\frac{1}{2}}\sum\limits_{n=0}^\infty\dfrac{(-1)^nx^{2n}}{(2n)!}dx=\int_0^\pi\sum\limits_{n=0}^\infty\dfrac{(-1)^nx^{2n+\frac{1}{2}}}{(2n)!}dx=\left[\sum\limits_{n=0}^\infty\dfrac{(-1)^nx^{2n+\frac{3}{2}}}{(2n)!\left(2n+\dfrac{3}{2}\right)}\right]_0^\pi=\sum\limits_{n=0}^\infty\dfrac{2(-1)^n\pi^{2n+\frac{3}{2}}}{(2n)!(4n+3)}$
Write x as a function of u: you have $u=\sin(x)$ so $x=\arcsin(u)$. It gets a little sticky when you have to simplify $\cos(\arcsin(u))$; remember the identity $\sin^2\theta+\cos^2\theta=1$, letting $\theta=\arcsin u$ (arcsin gives you an angle, recall) you get $\cos(\arcsin(u))=\sqrt{1-u^2}$.

Actually, this integral doesn't have a closed form; are you sure the problem wasn't $\int \sqrt{x} \cos^2 x dx$