# Computing $\int_0^1 \frac{\arcsin \sqrt x}{x^2-x+1} dx$ [duplicate]

Compute: $$\int_0^1 \frac{\arcsin \sqrt x}{x^2-x+1} dx$$

Answer: $$\frac{\pi^2}{6\sqrt 3}$$

My Attempt:

The obvious substitution: $$\arcsin \sqrt x=t$$. This transforms my integral (say $$I$$) to: $$I=\int_0^{\frac{\pi}{2}} \frac{t\cdot 2\sin t\cos t}{\sin^4-\sin^2+1}dt$$ Then the substitution $$t\rightarrow \left(\frac{\pi}{2}+0\right)-t$$. This leads to the denominator remaining the same.

Then, I added the "versions" of $$I$$ and on some simplifications, obtained: $$I=\frac{\pi}{4}\int_0^{\frac{\pi}{2}}\underbrace{\frac{\sin 2t}{1-\frac{\sin^2 2t}{4}}}_{f(t)}dt$$ Since $$f(x)=f\left(\frac{\pi}{2}-x\right)$$ we say: $$I=\frac{\pi}{2}\int_0^{\frac{\pi}{4}}\frac{\sin 2t}{1-\frac{\sin^2 2t}{4}}dt$$ Now the substitution $$2t=u$$ transforms it to: $$I=\frac{\pi}{4}\int_0^{\frac{\pi}{2}}\frac{\sin u}{1-\frac{\sin^2 u}{4}}du$$ Or, $$I=\pi\int_0^{\frac{\pi}{2}}\frac{\sin u}{4-\sin^2 u}du$$ Then I tried to decompose it into partial fractions and integrate them individually, by converting them in terms of $$\tan \frac u2$$ but the answer I'm getting doesn't match the given one.

• 2+sinu-2 should work for the last exp Aug 14, 2020 at 6:58
• You are almost done since $4- \sin^2x = 3+\cos^2x$ and substitute $\cos x =y$. Aug 14, 2020 at 7:22

Hint: Not all the substitutions work always. Also, you should first use $$t=0+1-x$$ then go for partial fraction $$I=\int_0^1 \frac{\sin^{-1} \sqrt x}{x^2-x+1} dx=\int_0^1 \frac{\sin^{-1} \sqrt {1-x}}{x^2-x+1}$$ Note that $$\sin^{-1}\sqrt{1-x}=\cos^{-1}\sqrt x$$.
$$\int_0^1 \frac{\arcsin\sqrt{x}}{\left(x-\frac{1}{2}\right)^2+\frac{3}{4}}dx = \frac{2}{\sqrt{3}}\arctan\left(\frac{2x-1}{\sqrt{3}}\right)\arcsin\sqrt{x}\Biggr|_0^1 - \frac{1}{\sqrt{3}}\int_0^1\frac{\arctan\left(\frac{2x-1}{\sqrt{3}}\right)}{\sqrt{x}\sqrt{1-x}}dx$$
$$= \frac{\pi^2}{6\sqrt{3}}-\frac{1}{\sqrt{3}}\int_{-\frac{1}{2}}^{\frac{1}{2}}\frac{\arctan\left(\frac{2x}{\sqrt{3}}\right)}{\sqrt{\frac{1}{2}+x}\sqrt{\frac{1}{2}-x}}dx = \frac{\pi^2}{6\sqrt{3}}$$
$$I=\pi\int_{0}^{\pi/2} \frac{\sin u}{4-\sin^2 u}= \pi \int_{0}^{\pi/2} \frac{\sin u}{3+\cos^2 u} du= \pi\int_{0}^{1} \frac{dv}{3+v^2}dv=\frac{\pi}{\sqrt{3}} \tan^{-1}(v/\sqrt{3})|_{0}^{1}= \frac{\pi^2}{6 \sqrt{3}}.$$ Lastly, we have used $$v=\cos u$$.