When can you manipulate the integral variable? Proof that if $Z$ is standard normal, then Z^2 is distributed Chi-Square (1).
For instance, can you always make $dz$ into $\frac{dw}{a}$ and then treat $\frac{1}{a}$ as a constant that you can take out of the integral?
 A: There are at least two ways of showing. 
Method 1: using the MGF (moment generating function) method (or the more general characteristic function method), as shown in the answer here.
Method 2: using the cumulative distribution function method.
First recall that the pdf for the Chi-Square with 1 df is 
$$f(z)=\frac{1}{\sqrt{2\pi}}\frac{1}{\sqrt{z}}e^{-\frac{z}{2}}, z>0\ \ \text{and}\ \ f(z)=0, z\le 0.$$
Prove: if $X\sim$ Normal(0,1) then $Y=X^2\sim$Chi-Square(1).
Let $F_Y(y)$ be the cdf (cumulative density function) for $Y$.
$$F_Y(y)=P(Y\le y)=P(X^2\le y)=P(-\sqrt{y}\le X \le \sqrt{y})$$
$$F_Y(y)=\int_{-\sqrt{y}}^{\sqrt{y}}\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}\ dx=\int_{0}^{\sqrt{y}}\frac{2}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}\ dx$$
The last development uses the fact that the Normal(0,1) pdf is a function symmetric around zero (an even function). 
Now making a change of variable in the integral, $x=\sqrt{z}$, with $dx = \frac{1}{2\sqrt{z}}dz$ leads to
$$F_Y(y)=\int_{0}^{{y}}\frac{2}{\sqrt{2\pi}}e^{-\frac{z}{2}}\frac{1}{2\sqrt{z}}\ dz=\int_{0}^{{y}}\frac{1}{\sqrt{2\pi}}\frac{1}{\sqrt{z}}e^{-\frac{z}{2}}\ dz=\int_{0}^{{y}}f(z)\ dz$$
The last step shows that the cdf for $Y$ is defined by the integral of a function $f(z)$ that is exactly the pdf for the Chi-Square with 1 df, therefore $Y$ must have that distribution.
