How to find indefinte integral $\int_{0}^{\sqrt\pi} 2x\cos(\frac{x^2}{4})dx$ using substitution of $u=\cos(\frac{x^2}{4})$? 
Can someone help me find $\int_{0}^{\sqrt\pi} 2x\cos(\frac{x^2}{4})dx$ using the substitution $u=\cos(\frac{x^2}{4})$ step-by-step?

All I've done so far is:
$$\frac{du}{dx}=-\frac{1}{2}x\sin(\frac{x^2}{4})$$
and
$$x=\sqrt\pi \longrightarrow u=\frac{\sqrt2}{2}$$
$$x=0 \longrightarrow u=1$$
$$\therefore \int_{1}^{\frac{\sqrt2}{2}}\frac{-4u}{\sqrt{1-u^2}}du$$
This is where I don't know how to continue, can ayone help me understand how to finish this off?

From your steps
\begin{align*}
&\int_{1}^{\frac{1}{\sqrt{2}}} \frac{-4u}{\sqrt{1 - u^2}}du\\
=&~2 \int_{0}^{\frac{1}{2}} \frac{dt}{\sqrt{t}} \quad \text{on the substituion } (1 - u^2) = t\\
=&~2 \int_{0}^{\frac{1}{\sqrt{2}}}\frac{w}{w}d\omega\quad \text{on the substituition } t = \omega^2\\
=&~4 \bigg[\frac{1}{\sqrt{2}}\bigg] = 2\sqrt{2} 
\end{align*}
