(I´m not an english native speaker so sorry if I don´t let my self be clear or write something wrong)
I was solving the last integral from this bprp video and got this answer:
$$\int_{}^{}\frac{x^2}{\sqrt{1-x^2}}dx=-\frac{1}{2}\left(\arccos(x)+\sqrt{(1-x^2)}x\right)+C$$
instead of replacing $x$ with $\sin(u)$, I replaced it with $\cos(u)$ to see how the answer will change. However, my answer is wrong according to WA
The thing is that my answer and WA´s answer only differs by a $+\pi/4$ in the end (if you see well both functions have the same "shape"). Matter fact if I add a $+\pi/4$ my answer is correct according to WA
(Answer from WA):
$$\int_{}^{}\frac{x^2}{\sqrt{1-x^2}}dx=-\frac{1}{2}\left(\arccos(x)+\sqrt{(1-x^2)}x\right)+\pi/4+C$$
Shouldn´t that pi/4 be the integration constant and therefore my first answer be correct? I have checked my procedure for almost an hour and I´m going crazy trying to find what I did wrong/skipped. What mindblows me the most is the fact that there is $\pi/4$ inside of the answer! I´m currently taking calc 2 and I´ve done a lot of integrals and i´ve never seen an integral that gives you a constant apart from the integration constant (apart from integral of 0 ofc). I also think that´s impossible but i don´t know how to prove it. Please help
PD1: One thing I did while solving the integral was considering $\sqrt{\cos{^2}(u)}=\cos(u)$ instead of $\left| \cos(u) \right|$ but I don´t think that causes the appereance of the $+\pi/4$ at the end.
PD2: The appereance of $+\pi/4$ surprises me because I can´t think of a function $f(x)$ such that $\int_{}^{}f(x) dx=F(x)+B+C$ where only $C$ is the integration constant.