Why is the derivative changed during trigonometric substitution? When using trigonometric substitution to find the indefinite integral of an expression, the derivative typically begins as $dx$. Once some expression is substituted for the $x$ in the expression, the derivative of the function is also changed. Why is that?
Examples of trigonometric substitution are here.
 A: It's just the chain rule:
$$
\int \underbrace{f'(g(x))}\ \underbrace{g'(x)\,dx} = \int f'(u)\,du = f(u)+C = f(g(x))+C  
$$
Postscript in response to comments:
The chain rule is differentiation by substitution.  Suppose $y = f(g(x))$.  Then do the substitution $u=g(x)$.  Then $y=f(u)$ and $$\frac{dy}{dx}=(f\circ g)'(x)=f'(u)g'(x).$$  Notice that one can write $$dy = f'(u)\ \underbrace{g'(x)\,dx} = f'(u)\,du.$$
That's the chain rule and that's differentiation by substitution.
A: when you use a transformation say $(x, f(x))$ to $(u, f(u)$ where $u = u(x)$ the domain of $f(u)$ is changed. 
for example let us take $f(x) = (2x)^3$ on $D = [0,1]$ and consider the transformation $x = u/2.$   the transformed function $f(x) = g(u) = u^3$ and the domain of $g$ is $[0,2].$ 
if you are finding the area $\int_0^1 (2x)^3 \ dx = 2$ which is the same as 
$\int_0^2 u^3 \frac{du}{2} = 2$
you can think of the transformation sends the rectangle $f(x)dx$ to $f(u/2) d(u/2)$
i mean by transformation of rectangles is this: draw a graph $y = 8x^3$ on $0 \le x \le 1.$  partition the interval $[0,1]$ into ten equal pieces. pick a piece say $[3/10, 4/10]$ that is the base of the rectangle. now you pick any point $c$ in this base and make a rectangle of height $f(c)$ by the transformation $u = 2x,$ we mean all the $x$ values are multiplied by two(magnified/stretched) and the $y$ values stay the same. the base $[3/10, 4/10]$ in the $xy$ will go to $6/10,8/10]$ the height of the rectangle stays put. to take care of the stretch we needed to divide by $2$ that is why you see that $du/2.$
i hope this helps. it is a good idea and to draw the graphs of $y = 8x^3$ and do a transformation $u = 2x.$
