Help needed in derivation related to chi square distribution (Basic query)

Question

Assuming $x$ and $y$ are two chi square random variables. The pdf of $x+y$ can be obtained following the steps provided in https://en.wikipedia.org/wiki/Proofs_related_to_chi-squared_distribution . However, I am unable to understand the part where the limit of $A$ is taken from $0$ to $\frac{B^2}{4}$ and the very next step where $A=\frac{B^2}{4}\sin^2(t)$ substitution is used. I will be very thankful to you for your help in making me understand how these limits are found and further how the substitution is done. Thanks in advance.

Answer 1

If $A = \frac{B^2}{4} \sin^2 (t)$, then

$$dA = \frac{B^2}{4} 2 \sin (t) \cos(t) dt = \frac{B^2}{2} \sin (t) \cos(t) dt$$

and $A = \frac{B^2}{4}$ implies $\sin^2(t) = 1$ and thus $t = \frac{\pi}{2}$ is the first solution.

Using the Euler identity $1 - \sin^2(t) = \cos^2(t)$, we also have

$$A^{-\frac{1}{2}} (B^2 - 4 A)^{-\frac{1}{2}} = \left(\frac{4}{B^2\sin^2(t)}\right)^\frac{1}{2} \left(\frac{1}{B^2(1-\sin^2(t))}\right)^\frac{1}{2} = \frac{2}{B\sin(t)}\frac{1}{B\cos(t)}$$

Hence the substitution yields

$$\int_0^{\frac{B^2}{4}} A^{-\frac{1}{2}} (B^2 - 4 A)^{-\frac{1}{2}} dA = \int_0^{\frac{\pi}{2}}\frac{2}{B\sin(t)}\frac{1}{B\cos(t)} \frac{B^2}{2}\sin (t) \cos(t) dt = \int_0^\frac{\pi}{2} dt$$