I am looking at an integral of the form
$$\int^{x_{\rm max}}_0 \frac{2 \sqrt{a} }{\sqrt{-a x^2 + b x + c}}dx$$
Where the upper limit is given by $x_{\rm max} = \frac{b + \sqrt{b^2 + 4 a c}}{2 a}$. Here $a, b$ and $c$ are all real and $a > 0$ etc. The answer apparently takes the form
\begin{align}
4 \arctan \left(\sqrt{ \frac{\sqrt{b^2 + 4 a c} + b}{\sqrt{b^2 + 4 a c} – b} } \right)
\end{align}
The Euler substitution, completing the square, trig substitutions, etc, do lead to solutions of the form $\arcsin(\cdot)$, and hence $\arctan(\cdot)$ via $$\arcsin(x) = \arctan\frac x {\sqrt{1-x^2}}$$
But, I can't seem to get something in the form above. This is a classic integral that can be found in any of the standard tables, e.g. 226 on Pg. 15 of https://www.math.stonybrook.edu/~bishop/classes/math126.F20/CRC_integrals.pdf. I'm probably doing something silly here. Thanks in advance for any insight here!
Best Answer
Apply the Euler substitution $$\sqrt{-a x^2 + b x + c}=\sqrt a\bigg(x-\frac{b-\sqrt{b^2+4ac}}{2a}\bigg)t$$ and note that the bounds become $(0,x_{max})\to (t_{max},0)$ with $$t_{max}=\sqrt{\frac{\sqrt{b^2+4ac}+b}{\sqrt{b^2+4ac}-b}}$$ Then
\begin{align} \int_0^{x_{max}}\frac{2\sqrt a}{\sqrt{-a x^2 + b x + c}}dx =\int^{t_{max}}_0\frac{4}{1+t^2}dt= 4 \tan^{-1} (t_{max}) \end{align}