# Help with Inverse Tangent Integral

Evaluate

$$\large \int _{ -1 }^{ 1 }{ \left( \cot^{-1}{ \frac { 1 }{ \sqrt { 1-{ x }^{ 2 } } } } \right) } \left( \cot^{-1}{ \frac { x }{ \sqrt { 1-{ \left( { x }^{ 2 } \right) }^{ \left| x \right| } } } } \right) dx\quad$$



Unfortunately, this time, I am unable to give any of my own inputs since I haven't been able to think of anything at all. I'm truly sorry for this; I would certainly have given all my inputs had I had the fortune to think of something relevant. The only thing which struck me was that the denominator in the first bracket of the integrand was the derivative of the Inverse Sine Function. I just cannot get any way of solving this and have been completely stumped with this. Would somebody please be so kind as to help me solve this? If possible, please could you refrain from using any special functions other than the Beta, Gamma and Digamma Functions? I would be extremely grateful for your assistance. Many, many thanks in advance!

EDIT: Please could you answer my comments to Robert Sir's solution? Thanks very much!

• Do you have any reason to think there is a closed form for this? It seems rather unlikely to me. – Robert Israel Jun 8 '15 at 20:28
• Sir, I think a closed form does exist (even though it seems improbable). The person who gave this said there was one. Unfortunately I won't be in contact with the aforementioned friend for a long time and have henced asked for guidance here. EDIT: Thanks so much Sir:) I've deleted mine too. – Ishan Jun 8 '15 at 20:31

## 1 Answer

Ah, there's a trick to it. Let your integral be $J = \int_{-1}^1 f(x)\; dx$. By symmetry, this is also $\int_{-1}^1 f(-x)\; dx$. Now note that $$f(x) + f(-x) = \dfrac{\pi^2}{2} - \pi \arctan\left(\dfrac{1}{\sqrt{1-x^2}}\right)$$ and this can easily be integrated, resulting in $$J = \dfrac{\pi^2}{2} \left(\sqrt{2}-1\right)$$

• Firstly Sir, I'm really, really thankful that you discovered a method. Hats off to you Sir, that too for thinking of this. I had completely forgotten the property that $$\int_a^b f(x)dx=\int_a^b f(a+b-x)dx$$ Sorry for 2 exceedingly stupid doubts, Sir. But I couldn't understand how you got $$f(x) + f(-x) = \dfrac{\pi^2}{2} - \pi \arctan\left(\dfrac{1}{\sqrt{1-x^2}}\right)$$, nor could I understand how to integrate $\pi \arctan(\dfrac{1}{\sqrt{1-x^2}})$ to get the result. – Ishan Jun 8 '15 at 20:49
• Sir, for the integral of $\pi \arctan(\dfrac{1}{\sqrt{1-x^2}})$, I made the substitution $x=\sin(\theta)$. However, that seems to result in integrating $\pi \arctan \bigg (|\sec(\theta)|\bigg )$ – Ishan Jun 8 '15 at 20:56
• Sir, are this is how far I could go:( $$I=\int _{ -1 }^{ 1 }{ \left( \cot^{-1}{ \frac { 1 }{ \sqrt { 1-{ x }^{ 2 } } } } \right) } \left( \cot^{-1}{ \frac { x }{ \sqrt { 1-{ \left( { x }^{ 2 } \right) }^{ \left| x \right| } } } } \right) dx\quad$$ $$I=\int _{ -1 }^{ 1 }{ \left( \cot^{-1}{ \frac { 1 }{ \sqrt { 1-{( -x) }^{ 2 } } } } \right) } \left( \cot^{-1}{ \frac { -x }{ \sqrt { 1-{ \left( {( -x) }^{ 2 } \right) }^{ \left| -x \right| } } } } \right) dx\quad$$ After this, we have to add both, but how does that get the result you wrote? I couldn't understand:( – Ishan Jun 8 '15 at 21:05
• Hint: $(-x)^2 = x^2$ and $|-x| = |x|$. How does $\cot^{-1}(-t)$ relate to $\cot^{-1}(t)$? – Robert Israel Jun 8 '15 at 22:19
• Sir, $\cot^{-1}(-t)=\pi - \cot^{-1}(t)$ So, $$2I=\int _{ -1 }^{ 1 }{ \left( \cot^{-1}{ \frac { 1 }{ \sqrt { 1-{x}^{ 2 } } } } \right) } \bigg ( \left( \cot^{-1}{ \frac { -x }{ \sqrt { 1-{ \left( {( -x) }^{ 2 } \right) }^{ \left| -x \right| } } } }\right)+\left( \cot^{-1}{ \frac { x }{ \sqrt { 1-{ \left( { x }^{ 2 } \right) }^{ \left| x \right| } } } } \right)\bigg ) dx\quad$$ Using $\cot^{-1}(-t)=\pi - \cot^{-1}(t)$, $$\Longrightarrow I=\dfrac{\pi}{2} \int_{-1}^1 \cot^{-1} \bigg (\frac { dx }{ \sqrt { 1-x^ 2} }\bigg)$$ But Sir, how would I integrate this? Please do help me Sir. – Ishan Jun 9 '15 at 5:34