Consdering, just as you apparently did, $$I=\int \frac{\sqrt{1-x}}{\sqrt{x}}\,dx$$ First $$x=u^2\implies dx=2u\,du\implies I=2\int \sqrt{1-u^2}\,du$$ Second $$u=\sin(t)\implies du=\cos(t)\,dt \implies I=2\int \cos^2(t) dt=\int (1+\cos(2t))\,dt$$ $$I=t+\frac{1}{2} \sin (2 t)+C$$ Now, back to $x$ $$t=\sin^{-1}(u)=\sin^{-1}(\sqrt x) \implies I=\sin ^{-1}(\sqrt x)+\frac 12\sin ^{-1}\left(2\sin ^{-1}(\sqrt x)\right)+C$$ which could simplify to $$I=\sqrt{(1-x) x}+\sin ^{-1}\left(\sqrt{x}\right)+C$$
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Begin with
$$A = \int \frac{\sec^2 \t}{\sin \t} \, \d \t$$
Use the Pythagorean identity $\sec^2 \t = 1 + \tan^2 \t$. Then with some algebra,
$$A = \underbrace{\int \csc \t \, \d \t}_{\ds =: \I_1} + \underbrace{\int \frac{\sin \t}{\cos^2 \t} \, \d \t}_{\ds =: \I_2}$$
$\I_2$ is trivial enough with the $u$-substitution $u := \cos \t$, which would give $\d u = - \sin \t \, \d \t$, so
$$\I_2 = - \int u^{-2} \, \d u = \frac{1}{u} + C = \frac{1}{\cos \t} +C = \sec \t + C$$
The first is a bit less trivial, but if you're familiar with the technique for solving $\int \sec(x) \, \d x$, it's quite similar. We have
$$\I_1 = \int \csc \t \frac{\csc \t + \cot \t}{\csc \t + \cot \t} \, \d \t$$
Use $u := \csc \t + \cot \t$; then $\d u = (- \csc^2 \t - \csc\t \cot \t) \, \d \t$, i.e. $-1$ times the numerator:
$$\I_1 = - \int \frac{1}{u} \, \d u = - \ln |u| + C = - \ln \big| \csc \t + \cot \t \big| + C$$
Therefore,
$$A = \sec \t - \ln \big| \csc \t + \cot \t \big| + C$$
Note that you had $\t = \arctan(x/\ell)$ and you can finish from here; what remains is basic trig and algebra.
Best Answer
Your answer and WA's are equivalent.
$\arccos \dfrac x5 + \arcsin \dfrac x5 = \dfrac {\pi}2$, so $\arcsin \dfrac x5 = \dfrac {\pi} 2 - \arccos \dfrac x5$,
so $\arcsin \dfrac x5 + C_1 = -\arccos \dfrac x5 + C_2$, where $C_1-C_2=\dfrac\pi2$.