$\int \frac{x^2}{\sqrt{5-x^2}}dx$ using substitution $u = \sqrt{5-x^2}$? I was trying to solve $$\int\frac{x^2}{\sqrt{5-4x^2 }} dx$$ by using the substitution
$$
u=\sqrt{5-4x^2}\\
dx = \frac{\sqrt{5-4x^2}}{-4x}du$$ So the integral can be written as $$\int \left(-\frac{x}{4}\right) du$$ and
$$
x= \begin{cases}
\sqrt{(5-u^2)/4}, & x >0\\
-\sqrt{(5-u^2)/4}, &x<0
\end{cases}
$$
I've rewritten the integral as
$$-\frac18\int\sqrt{5-u^2}du$$
using trig substitution with $$u= \sqrt5\sin(t)$$ we have
$$-\frac5{16}t - \frac5{16} \sin(t)\cos(t)$$ $$t=\arcsin\left(\frac{u^2}{\sqrt5}\right)$$
If we now reverse the substitutions and simplify, I got the final answer as $$-\frac5{16} \arcsin\left(\sqrt{\frac{5-4x^2}5}\right) - \frac{x}8\sqrt{5-4x^2} $$However this differs from the answer in my textbook which is:
$$-\frac{x}{8}\sqrt{5-4x^2}+\frac5{16}\arcsin\left(\frac{2x}{\sqrt5}\right)+C$$
My textbook used different methods that are simpler and easier, but I want to know where I made a mistake.

Note that
$$\arcsin\sqrt{\frac{5-4x^2}5}= \arccos\sqrt{1-\frac{5-4x^2}5}
=\arccos\frac{2x}{\sqrt5}=\frac\pi2-\arcsin\frac{2x}{\sqrt5}
$$
Therefore, the two results differs by the constant $-\frac\pi2$.
