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If we substitute $cos^{-1}(x)$ with $z$ then,corresponding values of $x=1$ are $z=0,2\pi,4\pi...$ And correspond values of $x=0$ are $z=\pi/2,3\pi/2,5\pi/2...$

Now if for $x$ belongs to $[0,1]$ corresponding value is taken $z$ belongs to $[\pi/2,0]$ then the value of integral comes out to be $(\pi^2)/8$ but for $z$ belongs to $[5\pi/2,2\pi]$ the value of integral comes out to be $9(\pi^2)/8$

Which interval should be taken?

I apologise for stating anything wrong if I have done any.

Edit: previously the question was written about {cos^-1(x)}^2 by mistake

MSKB
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  • Is $\cos^{-1}$ the inverse function of $\cos$ or $\frac{1}{\cos}$? –  Mar 10 '21 at 16:21
  • $\arccos$ is by definition the inverse of the function $\left.\cos\right\rvert_{[0,\pi]}:[0,\pi]\to[-1,1]$ and usually the preferred section of $\cos$ in the real variable is that one. –  Mar 10 '21 at 16:29
  • I think, if we consider $y=\frac{cos^{-1}(x)^2}{\sqrt{1-x^2}}$, then $y$ takes only non-negative values and thus, we will consider the first quadrant. Thus the former interval must be considered. – Cherryblossoms Mar 10 '21 at 16:30
  • Do we have to consider principle values [email protected] – MSKB Mar 10 '21 at 16:44

2 Answers2

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As @Gae.S. notes, you should take $\arccos0=\pi/2,\,\arccos1=0$ so$$\int_0^1\frac{\arccos^2xdx}{\sqrt{1-x^2}}=\left[-\frac13\arccos^3x\right]_0^1=\frac{\pi^3}{24}.$$

J.G.
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$$I=\int_{0}^{1} \frac{(\cos^{-1}x)^2}{\sqrt{1-x^2}} dx$$ Let $\cos^{-1} x=t \implies -\frac{dx}{\sqrt{1-x^2}} =dt$ So $$=\int_{0}^{\pi/2} t^2 dt= \frac{t^3}{3}|_{0}^{\pi/2}=\frac{\pi^3}{24}.$$

Z Ahmed
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