Intuition behind the substitution method of integration I want to know what is the rationale behind substitution method of integration.I'm very familiar with the following sort of integration but I don't understand why we substitute 
i.e use one variable in terms of another variable.
For the integral, $\int 2x\sqrt{x^2+3}\,dx$ we can use the substitution method in the following manner.
If $z=x^2+3,$ then $dz=2xdx$. We have $\int\sqrt{z}dz$.
My question is that why $\int\sqrt{z}dz$=$\int 2x\sqrt{x^2+3}\,dx$? Is there rationale behind it?
 A: Substitution can seem intuitive when approached from the perspective of the differential. Rather than the formal definition of the differential, here are a couple of examples that make the meaning pretty clear:
$ d \ (x^3 + a) = 3x^2dx $, and similarly $ d \ (\sin x) = \cos x \ dx $, and so on.
In other words:
\begin{equation*}
d \ [f(x)] = \frac{df}{dx} \cdot dx
\end{equation*}
Now if you think of the $dx$ or $dz$ term that appears in an integral in the same way, you will notice that this term defines the so called "variable of integration".
In particular, considering the example of the integral given in the question, notice first that:
$ d\, (x^2 + 3) = 2x\, dx $
And therefore, the suggested integral can be rewritten as:
\begin{eqnarray*}
 \int 2x \ \sqrt{x^2 + 3} \ \ dx &=&  \int \sqrt{x^2 + 3} \ \ (2x\, dx) \\
 &=& \int \sqrt{x^2 + 3} \ \ d\, (x^2 + 3) \ \ \ \ \ \mbox{[using the differential noted above]}
\end{eqnarray*}
The integral now looks like it is of the form:
\begin{equation*}
\int \sqrt{z} \ dz
\end{equation*}
The process of substitution is effectively equivalent to the above process.
A: Let $f : [a,b] \rightarrow \mathbb{R}$ be continous, let $F$ be its antiderivative and let $g : [\alpha,\beta] \rightarrow [a,b]$ be differentiable, strictly nondecreasing andso that we have $g(\alpha) = a, g(\beta) = b$. Under these assumptions, due to the fundamental theorem of calculus we have
$$F(b) - F(a) = \int_{a}^{b} f(x) dx$$ 
but also
$$F(g(\beta)) - F(g(\alpha)) = \int_{\alpha}^{\beta} (F \circ g)' dx = \int_{\alpha}^{\beta} F'(g(x)) \cdot g'(x) dx = \int_{\alpha}^{\beta}f(g(x))\cdot g'(x) $$
As $g(\alpha) = a, g(\beta) = b$, we conclude
$$ \int_{a}^{b} f(x) dx = \int_{\alpha}^{\beta}f(g(x))\cdot g'(x) $$
