Changing the interval of definite integral I have to find this integral: $$\int_{-1}^{1} \sqrt {1-x^2}dx$$
To solve this, I use one of Euler's substitutions: $\sqrt{1-x^2} = t(x-1)$, from this: $x = $$ {t^2 - 1}\over {t^2 + 1}$, $dx =$$ {4t}\over{t^2+1}$$dt$. The thing is that I don't know how to transformate interval: what are the $a,b $ in $\int_{a}^{b}$$ {4t^2}\over{t^2+1}$$dt$?
 A: Since you discovered that $x = \frac{t^2-1}{t^2+1}$, you just substitute $1$ and $-1$ on this formula and find the $t$ that solves it.
(Edit: Since there is no $t$ such that $\frac{t^2-1}{t^2+1}=1$, then you can't substitute this way, because to substitute, you must find a bijective function $f:[a',b'] \to [a,b]$ such that $\ f(a)=a'$ and $\ f(b)=b'$ or $\ f(a)=b'$ and $\ f(b)=a'$. So, you substitute $x$ for $f(u)$ and $\mathrm{d}x = f'(u)\mathrm{d}u$ and everything goes fine.)
But there is a more simple way to solve this integral. If you look close, you will notice that the graph of this function is the upper half of the circunference $x^2+y^2=1$, since $y=\sqrt{1-x^2} \implies y^2+x^2=1$.
This way, you know that this integral is the area of the half circunference, wich is $\pi/2$.
So, you are in the upper half circle. Remembering the trigonometric circle, you substitute $x=\cos{u}$, wich gives $\mathrm{d}x=-\sin{u}\mathrm{d}u$. To find the integrating interval, you substitute $x$ for $1$ and $-1$, giving $x=1 \implies u=0$ and $x=-1 \implies u=\pi$, wich are the exact values you were looking for, since they are the angles that define the upper half circle!
The integral ends up being:
$\int^{0}_{\pi}-\sin^2{u}\mathrm{d}u = \int^{\pi}_{0}\frac{1-\cos{2u}}{2}\mathrm{d}u=\frac{\pi}{2}$
Wich is just what we wanted. Remember that $\cos{2x}=\cos^2{x}-\sin^2{x}$ to find that $\sin^2{x} = \frac{1-\cos{2x}}{2}$ and the rest is easy.
(Edit 2: I overlooked a little mistake on the integral part, but I corrected it now.)
A: Hint:
For the lower bound, when $x=-1$, what is $t$? 
For the upper bound, when $x=1$, what is $t$?
If $t$ is undefined for one of the bounds, then you must use another substitution or method to solve the integral. 
A: You have $\sqrt{1-x^2} = t(x - 1) \implies t = \dfrac{\sqrt{1-x^2}}{x-1},\;x\neq 1$
So there's a problem when $x = 1$, though when $x = -1, t=0$. 

I'd suggest, instead, making the substitution $x = \sin\theta\implies dx = \cos \theta\,d\theta$. 
If $\sin \theta = -1 $, then pick $ \theta = -\pi/2$ as lower bound, and if $\sin \theta = 1,$ then pick $ \theta = \pi/2$ as the upper bound.
Then we have $$\int_{-\pi/2}^{\pi/2} \cos^2 \theta \,d\theta$$
Can you take it from here?
