Variable change in integral I need to calculate the integral $$\int_0^{2\pi}\sqrt{1-\cos\,x}dx$$ but when I try the substitution with $u = 1-\cos\,x$ both integration limits goes to 0.
The same happens using the t-substitution.
 A: You can't use that substitution over the entire interval $0 \leq x \leq 2\pi$, because $1-\cos x$ is not one-to-one there. Instead try splitting the integral into $\int_0^\pi \sqrt{1-\cos x}dx+\int_\pi^{2\pi} \sqrt{1-\cos x} dx$, and using your substitution on each one, separately.
A: Note that $\cos(2x)=\cos^{2}(x)-\sin^{2}(x)$
so $\vert\sin(x)\vert=\sqrt{\frac{1-\cos(2x)}{2}}$.
So redoing the calculation with $x$ as $\frac{x}{2}$ will give you 
$\vert\sin(\frac{x}{2})\vert=\frac{1}{\sqrt{2}}\sqrt{1-\cos(x)}$.
Now you can solve the integral by splitting it into $[0,\pi]$ and $[\pi,2\pi]$ and using substitutions.
A: Hint: Note that 
\begin{align}
1-\cos(x)
=
&
1-\cos\left(2\cdot \frac{x}{2}\right)
\\
=
&
1+\sin^2\left(\frac{x}{2}\right)-\cos^2\left(\frac{x}{2}\right)
\\
=
&
\sin^2\left(\frac{x}{2}\right)+\sin^2\left(\frac{x}{2}\right)
\\
=
&
2\sin^2\left(\frac{x}{2}\right)
\end{align}
implies 
$$
\int_{0}^{2\pi}\sqrt[2\,]{1+\cos(x)}\,\mathop{d}x=
2\cdot\int_{0}^{2\pi}\left|\sin\left(\frac{x}{2}\right)\right|\,\mathop{d}x
$$
