Question

If the value of the definite integral
$\underset{-1}{\overset{1}{\int }}{\cot }^{-1}\left(\frac{1}{\sqrt{1-x^2}}\right)\cdot {\cot }^{-1}\left(\frac{x}{\sqrt{1-(x^2{)}^{\left|x\right|}}}\right)dx=\frac{{\pi }^2(\sqrt{a}-\sqrt{b})}{\sqrt{c}},$ where $a,b,c\in N$ in their lowest form, then the value of $(a+b+c)$ is

Answer 1

Let $I=\underset{-1}{\overset{1}{\int }}{\cot }^{-1}\left(\frac{1}{\sqrt{1-x^2}}\right)\cdot {\cot }^{-1}\left(\frac{x}{\sqrt{1-(x^2{)}^{\left|x\right|}}}\right)dx\dots \left(1\right)$

By using $\underset{a}{\overset{b}{\int }}f\left(x\right)dx=\underset{a}{\overset{b}{\int }}f(a+b-x)dx,$ we get
$I=\underset{-1}{\overset{1}{\int }}{\cot }^{-1}\left(\frac{1}{\sqrt{1-x^2}}\right){\cot }^{-1}\left(\frac{-x}{\sqrt{1-(x^2{)}^{\left|x\right|}}}\right)\Rightarrow I=\underset{-1}{\overset{1}{\int }}{\cot }^{-1}\left(\frac{1}{\sqrt{1-x^2}}\right)(\pi -{\cot }^{-1}\frac{x}{\sqrt{1-(x^2{)}^{\left|x\right|}}})dx\dots \left(2\right)$

Adding $\left(1\right)$ and $\left(2\right),$ we get
$2I=\underset{-1}{\overset{1}{\int }}\pi {\cot }^{-1}\left(\frac{1}{\sqrt{1-x^2}}\right)dx\Rightarrow 2I=\underset{-1}{\overset{1}{\int }}\pi {\tan }^{-1}\sqrt{1-x^2}dx\Rightarrow 2I=2\pi \underset{0}{\overset{1}{\int }}{\tan }^{-1}\sqrt{1-x^2}dx$
(As ${\tan }^{-1}\sqrt{1-x^2}$ is even function)
$\Rightarrow I=\pi \underset{0}{\overset{1}{\int }}1\cdot {\tan }^{-1}\sqrt{1-x^2}dx$

By integration by parts,
$I=\pi [{[{\tan }^{-1}\sqrt{1-x^2}\cdot x]}_0^1-\underset{0}{\overset{1}{\int }}\frac{x}{1+1-x^2}\cdot \frac{-x}{\sqrt{1-x^2}}dx]$
$\Rightarrow I=\pi \underset{0}{\overset{1}{\int }}\frac{x^2}{(2-x^2)\sqrt{1-x^2}}dx$

Let, $x=\sin \theta \Rightarrow dx=\cos \theta d\theta$
$\Rightarrow I=\pi \underset{0}{\overset{\pi /2}{\int }}\frac{{\sin }^2\theta }{(2-{\sin }^2\theta )}d\theta$
$\Rightarrow I=-\pi \underset{0}{\overset{\pi /2}{\int }}\frac{2-{\sin }^2\theta -2}{2-{\sin }^2\theta }d\theta$
$\therefore I=2\pi \underset{0}{\overset{\pi /2}{\int }}\frac{d\theta }{2-{\sin }^2\theta }-\frac{{\pi }^2}{2}$
$\Rightarrow I=2\pi \underset{0}{\overset{\pi /2}{\int }}\frac{{\sec }^2\theta d\theta }{2+2{\tan }^2\theta -{\tan }^2\theta }-\frac{{\pi }^2}{2}$
$\Rightarrow I=2\pi \underset{0}{\overset{\pi /2}{\int }}\frac{{\sec }^2\theta d\theta }{2+{\tan }^2\theta }-\frac{{\pi }^2}{2}$

Putting $\tan \theta =t$,
$I=2\pi \underset{0}{\overset{\infty }{\int }}\frac{dt}{2+t^2}-\frac{{\pi }^2}{2}$
$\Rightarrow I=2\pi {\left[\frac{1}{\sqrt{2}}{\tan }^{-1}\frac{t}{\sqrt{2}}\right]}_0^{\infty }-\frac{{\pi }^2}{2}$
$\Rightarrow I=\frac{{\pi }^2}{\sqrt{2}}-\frac{{\pi }^2}{2}=\frac{{\pi }^2(\sqrt{2}-1)}{2}$
Now, $I=\frac{{\pi }^2(\sqrt{a}-\sqrt{b})}{\sqrt{c}}\text{(given)}$
$\Rightarrow a=2,b=1$ and $c=4$
$\Rightarrow a+b+c=7$