Integrate $\sqrt{x^2 - 2x}$ $$\int{\sqrt{x^2 - 2x}}$$
I think I should be doing trig substitution, but which? I completed the square giving 
$$\int{\sqrt{(x-1)^2 -1}}$$
But the closest I found is  for
$$\frac{1}{\sqrt{a^2 - (x+b)^2}}$$ 
So I must add a $-$, but how? 
 A: The standard way to solve these problems is indeed with trigonometric substitutions. 
But it is not the only way to solve these. 
Another way that was used more before is called Euler substitutions

if we are to integrate $\sqrt{ax^2+bx+c}$ that has real roots $\alpha$ and
  $\beta$, then we can use the substitution $\sqrt{ax^2+bx+c}=x \cdot t$

Now, in your problem we have $\alpha=0$ and $\beta=2$, so we may choose to use the substitution 
$$ \sqrt{x^2-2x} = (x-0)\cdot t \ \Rightarrow \ x = \frac{2}{1-t^2}$$
Since this is homework and I am lazy, so from here I will only outline the details
$$ \begin{align} I & = \int \sqrt{x^2-2x}\,\mathrm{d}x \\
                   & = \int \frac{8t^2}{1-3t^2+3t^4-t^6}\,\mathrm{d}t \\
                   & = \int -\frac{8t^2}{(t+1)^3(t-1)^3}\,\mathrm{d}t \\
                   & = \int - \frac{1}{2}\frac{1}{t+1} + \frac{1}{(t+1)^3} 
- \frac{1}{2}\frac{1}{(t+1)^2} + \frac{1}{2}\frac{1}{(t-1)^2} - \frac{1}{(t-1)^3} - \frac{1}{2}\frac{1}{(t-1)^2} \, \mathrm{d}t \\
\end{align} $$ 
and from here the rest is obvious =)
A: In hopes of taking advantage of the identity
$$\tan^2\theta=\sec^2\theta-1,$$ make the substitution:
 $$x-1=\sec\theta, \quad dx=\sec\theta\,\tan\theta\,d\theta.$$ 
Then we have
$$
\int \sqrt{(x-1)^2-1}\, dx = \int\sqrt{\sec^2\theta -1}\, \sec\theta\tan\theta\,d\theta
=\int\tan^2\theta\sec\theta\,d\theta=\int (\sec^3\theta-\sec\theta)\,d\theta.
$$

To integrate $\sec\theta$, you can use a trick:
$$\tag{1}
\int\sec\theta\,d\theta
=\int\sec\theta \cdot {\textstyle{\sec\theta+\tan\theta\over\sec\theta+\tan\theta}}\,d\theta
=\int{ {\textstyle{\sec^2\theta+\sec\theta\tan\theta\over \sec\theta+\tan\theta}}}d\theta=\ln|\sec\theta+\tan\theta|+C\ \ \ \ \ \ \ 
$$
To integrate $\sec^3\theta$, you can start with integration by parts:
$$\eqalign{
\color{maroon}{\int\sec^3\theta\,d\theta}=\int\underbrace{\sec \theta}_u \,\underbrace{\sec^2\theta d\theta}_{dv}
&= \underbrace{\sec \theta}_u \,\underbrace{\tan\theta}_{v} -
\int\underbrace{\tan \theta}_v\, \underbrace{\sec \theta\tan\theta\, d\theta}_{du}\cr
&=\sec\theta\tan\theta -\int( \sec^3\theta-\sec\theta)\,d\theta\cr
&=\sec\theta\tan\theta \color{maroon}{-\int \sec^3\theta\,d\theta}+\int \sec\theta\,d\theta.
}
$$
Then, in the above,  solve for $\color{maroon}{\int\sec^3\theta\,d\theta}$:
$$\tag{2}
\int\sec^3\theta\,d\theta={1\over2}\sec\theta\tan\theta+{1\over2}\int\sec\theta\,d\theta
={1\over2}\sec\theta\tan\theta+{1\over2}\ln|\sec\theta+\tan\theta|+C.
$$
So, using $(1)$ and $(2)$:
$$
\int(\sec^3\theta-\sec\theta)\,d\theta 
={1\over2}\sec\theta\tan\theta -{1\over2}\ln|\sec\theta+\tan\theta|+C.
$$
Finally put everything back in terms of $x$ using $\theta=\sec^{-1}(x-1)$ (which I'll leave to you).
A: The main obstacle here is the square root.  It is likely that eliminating it will allow us to proceed.  So we want something squared minus 1 is the square of something.  That leaves secant (or cosecant) as the best option.
As N3buchadnezzar states, it is not the only option.  Consider the formulas
$(m+n)^2=m^2+2mn+n^2,(m-n)^2=m^2-2mn+n^2$
It is possible to do a substitution such that 
$(m+n)^2-1=(m-n)^2$
$(m+n)^2-(m-n)^2=1$
$4mn=1$
So any substitution $x-1=\frac{f(t)}2+\frac1{2f(t)}$ will work, though back substitution may be messy and will likely require the same properties to undo.  For example, let's do
$x-1=\frac12(e^t+e^{-t}),dx=\frac12(e^t-e^{-t})dt$
$\int\sqrt{(x-1)^2-1}dx=\int\frac12(e^t-e^{-t})\sqrt{\frac14(e^{2t}+e^{-2t}+2)-1}dt$=
$\int\frac12(e^t-e^{-t})\sqrt{\frac14(e^{2t}+e^{-2t}+2-4)}dt=$
$\int\frac12(e^t-e^{-t})\sqrt{\frac14(e^{2t}+e^{-2t}-2)}dt=\int(\sqrt{\frac14(e^{2t}+e^{-2t}-2)})^2dt=$
$\frac14\int e^{2t}+e^{-2t}-2dt=\frac18e^{2t}-\frac18e^{-2t}-\frac t2+C$
Now the problem of back-substitution.  How do we get back a function in terms of x?  Well, we have
$x-1=\frac12(e^t+e^{-t}),\sqrt{(x-1)^2-1}=\frac12(e^t-e^{-t})$
$x-1+\sqrt{x^2-2x}=e^t,t=\ln(x-1+\sqrt{x^2-2x})$
That you'll need for the $\frac t2$ term.  For the rest, you'd want to use
$\frac18e^{2t}-\frac18e^{-2t}=\frac12(\frac{e^t+e^{-t}}2)(\frac{e^t-e^{-t}}2)=\frac12(x-1)\sqrt{x^2-2x}$
So, like I said, secant is the best option. :)
A: I dont know i am right or wrong but i can do this example without using trigonometric substitution in following way,
\begin{align*}
\int\sqrt{x^{2}-2x}dx &=\int\sqrt{x^{2}-2x+1-1}dx\
                    &=\int\sqrt{(x-1)^{2}-1^{2}}dx\
                    &=\frac{x}{2}\sqrt{(x-1)^{2}-1}-\frac{1}{2}\log |x+\sqrt{(x-1)^{2}-1}|+c.
\end{align*}
I used the following formula of integration,
$\int\sqrt{x^{2}-a^{2}}dx=\frac{x}{2}\sqrt{x^{2}-a^{2}}-\frac{a^{2}}{2}\log |x+\sqrt{x^{2}-a^{2}}|+c.$
