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$\int_{0}^{1}\frac{\sin^{-1}\sqrt x}{x^2-x+1}dx$

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?

user1442
  • 1,212

5 Answers5

13

Let $$\displaystyle I = \int_{0}^{1}\frac{\sin^{-1}\sqrt{x}}{x^2-x+1}dx.............(1)$$

Now Replace $x\rightarrow (1-x)\;,$ We get

$$\displaystyle I = \int_{0}^{1}\frac{\sin^{-1}\sqrt{1-x}}{(1-x)^2-(1-x)+1}dx = \int_{0}^{1}\frac{\sin^{-1}\sqrt{1-x}}{x^2-x+1}dx$$

So we get $$\displaystyle I = \int_{0}^{1}\frac{\cos^{-1}\sqrt{x}}{x^2-x+1}dx...............(2)$$

Above we have use the formula $$\bullet\; \displaystyle \sin^{-1}(\sqrt{x})+\sin^{-1}\sqrt{1-x} = \sin^{-1}(\sqrt{x})+\cos^{-1}(\sqrt{x}) = \frac{\pi}{2}$$

Now Add these two equations, We get

$$\displaystyle 2I = \int_{0}^{1}\frac{\sin^{-1}\sqrt{x}+\cos^{-1}\sqrt{x}}{x^2-x+1} = \frac{\pi}{2}\int_{0}^{1}\frac{1}{x^2-x+1}dx = \frac{\pi}{2}\int_{0}^{1}\frac{1}{\left(x-\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2}dx$$

Now Put $\displaystyle \left(x-\frac{1}{2}\right) = t$ and $dx = dt$ and Changing Limt, We get

$$\displaystyle 2I = \int_{-\frac{1}{2}}^{\frac{1}{2}}\frac{1}{t^2+\left(\frac{\sqrt{3}}{2}\right)^2}dx = 2\cdot \frac{\pi}{2}\int_{0}^{\frac{1}{2}} \frac{1}{t^2+\left(\frac{\sqrt{3}}{2}\right)^2}dx = \frac{2\pi}{\sqrt{3}} \cdot \frac{\pi}{6} = \frac{\pi^2}{3\sqrt{3}}$$

So we get $$\displaystyle I = \int_{0}^{1}\frac{\sin^{-1}\sqrt{x}}{x^2-x+1}dx = \frac{\pi^2}{6\sqrt{3}}$$

juantheron
  • 53,015
10

$$ \begin{align} \int_0^1\frac{\sin^{-1}\left(\sqrt{x}\right)}{x^2-x+1}\,\mathrm{d}x &=\int_0^1\frac{\frac12\cos^{-1}(1-2x)}{x^2-x+1}\,\mathrm{d}x\tag{1}\\ &=\int_{-1}^1\frac{\cos^{-1}(x)}{3+x^2}\,\mathrm{d}x\tag{2}\\ &=\frac12\int_{-1}^1\frac{\cos^{-1}(x)+\cos^{-1}(-x)}{3+x^2}\,\mathrm{d}x\tag{3}\\ &=\frac\pi2\int_{-1}^1\frac1{3+x^2}\,\mathrm{d}x\tag{4}\\ &=\frac\pi{2\sqrt3}\int_{-1/\sqrt3}^{1/\sqrt3}\frac1{1+x^2}\,\mathrm{d}x\tag{5}\\ &=\frac\pi{\sqrt3}\tan^{-1}\left(\frac1{\sqrt3}\right)\tag{6}\\ &=\frac{\pi^2}{6\sqrt3}\tag{7} \end{align} $$ 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

robjohn
  • 345,667
5

Let, $$I=\int_{0}^{1}\frac{\sin^{-1}(\sqrt x)}{x^2-x+1}dx\tag 1$$ Now, using property of definite integral $\int_{a}^{b}f(x)dx=\int_{a}^{b}f(a+b-x)dx$ we get

$$I=\int_{0}^{1}\frac{\sin^{-1}(\sqrt {1-x})}{(1-x)^2-(1-x)+1}dx$$ $$I=\int_{0}^{1}\frac{\sin^{-1}(\sqrt {1-x})}{x^2-x+1}dx\tag 2$$ Now, adding (1) & (2), we get $$I+I=\int_{0}^{1}\frac{\sin^{-1}(\sqrt {x})}{x^2-x+1}dx+\int_{0}^{1}\frac{\sin^{-1}(\sqrt {1-x})}{x^2-x+1}dx$$ $$2I=\int_{0}^{1}\frac{\sin^{-1}(\sqrt {x})+\sin^{-1}(\sqrt {1-x})}{x^2-x+1}dx$$

$$I=\frac{1}{2}\int_{0}^{1}\frac{\sin^{-1}(\sqrt {x}\sqrt{1-1+x}+\sqrt{1-x}\sqrt{1-x})}{x^2-x+1}dx$$ $$=\frac{1}{2}\int_{0}^{1}\frac{\sin^{-1}(x+(1-x)}{x^2-x+1}dx$$ $$=\frac{1}{2}\int_{0}^{1}\frac{\sin^{-1}(1)}{x^2-x+1}dx$$ $$=\frac{1}{2}\int_{0}^{1}\frac{\frac{\pi}{2}}{x^2-x+1}dx$$ $$=\frac{\pi}{4}\int_{0}^{1}\frac{dx}{x^2-x+1}$$ $$=\frac{\pi}{4}\int_{0}^{1}\frac{dx}{\left(x-\frac{1}{2}\right)^2+1-\frac{1}{4}}$$ $$=\frac{\pi}{4}\int_{0}^{1}\frac{dx}{\left(x-\frac{1}{2}\right)^2+\left(\frac{\sqrt 3}{2}\right)^2}$$

$$=\frac{\pi}{4}\left[\frac{2}{\sqrt 3}\tan^{-1}\left(\frac{\left(x-\frac{1}{2}\right)}{\frac{\sqrt 3}{2}}\right)\right]_{0}^{1}$$ $$=\frac{\pi}{2\sqrt 3}\left[\frac{\pi}{6}+\frac{\pi}{6}\right]$$ $$=\frac{\pi^2}{6\sqrt 3}$$

Harish Chandra Rajpoot
  • 37,450
2

OP’s substitution works well $$ \begin{aligned} I & =2 \int_0^{\frac{\pi}{2}} \frac{\theta \sin \theta \cos \theta d \theta}{\sin ^4 \theta-\sin ^2 \theta+1} \\ & =2 \int_0^{\frac{\pi}{2}} \frac{\left(\frac{\pi}{2}-\theta\right) \cos \theta \sin \theta}{\cos ^4 \theta-\cos ^2 \theta+1} d \theta, \quad \textrm{ via } x\mapsto \frac\pi 2-x \\ & =\pi \int_0^{\frac{\pi}{2}} \frac{\cos \theta \sin \theta}{\cos ^4 \theta-\cos ^2 \theta+1} d \theta-2 \int_0^{\frac{\pi}{2}} \frac{\theta \sin \theta \cos \theta}{\cos ^4 \theta-\cos ^2 \theta+1} d \theta \end{aligned} $$ Fortunately, $\cos ^4 \theta-\cos ^2 \theta+1 = \sin ^4 \theta-\sin ^2 \theta+1$ and hence $$ I=\frac{\pi}{2} \int_0^{\frac{\pi}{2}} \frac{\cos \theta \sin \theta}{\cos ^4 \theta-\cos ^2 \theta+1} d \theta $$ Putting $c=\cos^2\theta$ further transforms the integral $$ \begin{aligned} I & =\frac{\pi}{4} \int_0^1 \frac{d c}{c^2-c+1} \\ & =\pi \int_0^1 \frac{d c}{(2 c-1)^2+3} \\ & =\frac{\pi}{2 \sqrt{3}}\left[\tan ^{-1}\left(\frac{2 c-1}{\sqrt{3}}\right)\right]_0^1\\&= \frac{\pi^2}{6 \sqrt{3}} \end{aligned} $$

Lai
  • 20,421
1

Integrate by parts, then substitute $x\mapsto\dfrac12-x$ in the subsequent integral to see that its contribution vanishes.

$$\begin{align*} \int_0^1 \frac{\sin^{-1} \sqrt x}{x^2 - x + 1} \, dx &= \frac{\pi^2}{6\sqrt3} - \frac1{\sqrt3} \int_0^1 \frac{\tan^{-1}\left(\frac{2x-1}{\sqrt3}\right)}{\sqrt x \sqrt{1-x}} \, dx \\ &= \frac{\pi^2}{6\sqrt3} - \frac2{\sqrt3} \underbrace{\int_{-\tfrac12}^{\tfrac12} \frac{\tan^{-1}\frac{2x}{\sqrt3}}{\sqrt{1-4x^2}} \, dx}_{=0} = \boxed{\frac{\pi^2}{6\sqrt3}} \end{align*}$$

user170231
  • 19,334