TITLE: $\int_{0}^{1}\frac{\sin^{-1}\sqrt x}{x^2-x+1}dx$
QUESTION [6 upvotes]: $\int_{0}^{1}\frac{\sin^{-1}\sqrt x}{x^2-x+1}dx$
Put $x=\sin^2\theta,dx=2\sin \theta \cos \theta d\theta$
$\int_{0}^{\pi/2}\frac{\theta.2\sin \theta \cos \theta d\theta}{\sin^4\theta-\sin^2\theta+1}$ but this seems not integrable.Is this a wrong way?Is there a different way to solve it?

REPLY [6 votes]: $$
\begin{align}
\int_0^1\frac{\sin^{-1}\left(\sqrt{x}\right)}{x^2-x+1}\,\mathrm{d}x
&=\int_0^1\frac{\frac12\cos^{-1}(1-2x)}{x^2-x+1}\,\mathrm{d}x\tag{1}\\
&=\int_{-1}^1\frac{\cos^{-1}(x)}{3+x^2}\,\mathrm{d}x\tag{2}\\
&=\frac12\int_{-1}^1\frac{\cos^{-1}(x)+\cos^{-1}(-x)}{3+x^2}\,\mathrm{d}x\tag{3}\\
&=\frac\pi2\int_{-1}^1\frac1{3+x^2}\,\mathrm{d}x\tag{4}\\
&=\frac\pi{2\sqrt3}\int_{-1/\sqrt3}^{1/\sqrt3}\frac1{1+x^2}\,\mathrm{d}x\tag{5}\\
&=\frac\pi{\sqrt3}\tan^{-1}\left(\frac1{\sqrt3}\right)\tag{6}\\
&=\frac{\pi^2}{6\sqrt3}\tag{7}
\end{align}
$$
Explanation:
$(1)$: $\sin^{-1}\left(\sqrt{x}\right)=\frac12\cos^{-1}(1-2x)$
$(2)$: substitute $x\mapsto\frac{1-x}2$
$(3)$: $\int_{-1}^1f(x)\,\mathrm{d}x=\frac12\int_{-1}^1(f(x)+f(-x))\,\mathrm{d}x$
$(4)$: $\cos^{-1}(x)+\cos^{-1}(-x)=\pi$
$(5)$: substitute $x\mapsto\sqrt3x$
$(6)$: integrate
$(7)$: simplify