Evaluate $\int_{-1}^{1} \cot^{-1} \left(\frac{1}{\sqrt{1-x^2}}\right) \cot^{-1}\left(\raise{3pt}\frac{x}{\sqrt{1-(x^2)^{|x|}}}\right)$

Question

$$\int_{-1}^{1} \left(\cot^{-1} \dfrac{1}{\sqrt{1-x^2}}\right) \left(\cot^{-1}\dfrac{x}{\sqrt{1-(x^2)^{|x|}}}\right)= \dfrac{\pi^2(\sqrt a-\sqrt b )}{\sqrt c}$$ , where a,b, c are natural numbers and are there in their lowest form, then find the value of a+b+c.

Using $\int_a^b f(x)dx= \int_a^bf(a+b-x{) dx}$, I got:

$$2I =2\pi\int_{0}^1 \cot^{-1}\left(\dfrac{1}{\sqrt{1-x^2}}\right) dx$$

Then, letting $x = \sin \theta$

$I = \pi\displaystyle\int_{0}^{\pi/2}\arctan (\cos\theta) \cos \theta d\theta$

After this I tried integration by parts but it gets really complicated with that? How do I continue?

EDIT: Please note that arccot(x) + arccot(-x)= $\pi$ $\ne 0$

Principal range of $\cot^{-1}x$ considered in the question is $(0,\pi)$

Answer 1

HINT:

Your integrand is an odd funtion, so:

$$\int_{-\text{n}}^\text{n}\text{y}\left(x\right)\space\text{d}x=0\tag1$$

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Using Mathematica I got:

Answer 2

I usually see $\cot^{-1}(-x)=-\cot^{-1}(x)$, but working with what is stated in the question, $$ \begin{align} &\int_{-1}^1\cot^{-1}\left(\frac1{\sqrt{1-x^2}}\right)\cot^{-1}\left(\raise{3pt}\frac{x}{\sqrt{1-\left(x^2\right)^{|x|}}}\right)\,\mathrm{d}x\tag1\\ &=\int_0^1\cot^{-1}\left(\frac1{\sqrt{1-x^2}}\right)\cot^{-1}\left(\raise{3pt}\frac{x}{\sqrt{1-\left(x^2\right)^{|x|}}}\right)\,\mathrm{d}x\\ &+\int_0^1\cot^{-1}\left(\frac1{\sqrt{1-x^2}}\right)\left[\pi-\cot^{-1}\left(\raise{3pt}\frac{x}{\sqrt{1-\left(x^2\right)^{|x|}}}\right)\right]\,\mathrm{d}x\tag2\\ &=\pi\int_0^1\tan^{-1}\left(\sqrt{1-x^2}\right)\,\mathrm{d}x\tag3\\ &=\pi\int_0^1\frac{\sqrt{1-x^2}}{1+x^2}\,\mathrm{d}x\tag4\\ &=\pi\int_0^{\pi/2}\frac{\cos^2(x)}{1+\sin^2(x)}\,\mathrm{d}x\tag5\\ &=\pi\int_0^{\pi/2}\frac2{1+\sin^2(x)}\,\mathrm{d}x-\frac{\pi^2}2\tag6\\ &=\pi\int_0^{\pi/2}\frac{2\sec^2(x)}{1+2\tan^2(x)}\,\mathrm{d}x-\frac{\pi^2}2\tag7\\ &=\pi\sqrt2\int_0^\infty\frac1{1+u^2}\,\mathrm{d}u-\frac{\pi^2}2\tag8\\ &=\pi^2\frac{\sqrt2-1}2\tag9 \end{align} $$ Explanation:
$(2)$: use $\cot^{-1}(-x)=\pi-\cot^{-1}(x)$ and substitute $x\mapsto-x$ on $[-1,0]$
$(3)$: $\cot^{-1}(x)=\tan^{-1}\left(\frac1x\right)$
$(4)$: substitute $x\mapsto\sqrt{1-x^2}$ then integrate by parts
$(5)$: substitute $x\mapsto\sin(x)$
$(6)$: add $1$ to the integrand and subtract $\frac{\pi^2}2$
$(7)$: multiply numerator and denominator of the integrand by $\sec^2(x)$
$(8)$: substitute $u=\sqrt2\tan(x)$
$(9)$: substitute $u=\tan(\theta)$ and integrate

Answer 3

Using the property $$\int_a^b f(x)=\int_a^b f(a+b-x)$$

The integral changes to $$I=\pi\int_0^1 \arctan \sqrt{1-x^2} dx$$

You might know the property that $$\int_a^b f(x)+\int_{f(a)}^{f(b)} f^{-1}(x) dx=-af(a)+bf(b)$$

Let $$J=\int_0^1 \arctan \sqrt{1-x^2} dx+\int_{\frac {\pi}{4}}^0 \sqrt {1-\tan ^2x } dx$$ Using above property $J=0$ Hence $$\pi\int_0^1 \arctan \sqrt{1-x^2} dx=\pi\int_0^{\frac {\pi}{4}} \sqrt {1-\tan ^2x } dx=I$$

Using the substitution $\tan x=\sin \theta$ in the right integral we get $$I=\pi\int_0^{\frac {\pi}{2}} \frac {\cos ^2\theta}{1+\sin^2\theta} d\theta=\pi\int_0^{\frac {\pi}{2}}\left( -1+\frac {2\sec^2\theta}{1+2\tan^2\theta}\right) d\theta$$

Hope you can continue further

Answer 4

your first line is wrong. See there:

http://www.wolframalpha.com/input/?i=%5Bcot%5E(-1)(x%2Fsqrt(1-(x%5E2)%5E%7Cx%7C))%5D+%2B+cot%5E(-1)(-x%2Fsqrt(1-(x%5E2)%5E%7Cx%7C))

Anyway, for $I = \pi\displaystyle\int_{0}^{\pi/2}\arctan (\cos\theta) \cos \theta\, \mathrm{d}\theta$ integration by parts would be good start. Let $f'=\cos \theta$ and $g=\arctan(\cos \theta)$. Try to continue.