$\int \frac{x^2}{\sqrt{5-x^2}}dx$ using substitution $u = \sqrt{5-x^2}$?

Question

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.

Answer 1

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 $-\frac\pi2$.

Answer 2

Integration by parts $$ \begin{aligned} I&=\int \frac{x^2}{\sqrt{5-x^2}} d x \\ & =-\int x d\left(\sqrt{5-x^2}\right) \\ & =-x \sqrt{5-x^2}+\int \sqrt{5-x^2} d x \\ & =-x \sqrt{5-x^2}+\int \frac{5-x^2}{\sqrt{5-x^2}} d x \\ & =-x \sqrt{5-x^2}+5 \int \frac{d x}{\sqrt{5-x^2}}-I \\ & =\frac{1}{2}\left[5 \sin ^{-1}\left(\frac{x}{\sqrt{5}}\right)-x \sqrt{5-x^2}\right]+C \end{aligned} $$