Exact value of $\int_0^{\frac\pi 4}\operatorname{arccot}\sqrt{2-\sec^2\theta}\,\mathrm d\theta$

Question

To find the exact value of a difficult triple integral, I need to evaluate the definite integral below: $$\int_0^{\frac\pi 4}\operatorname{arccot}\sqrt{2-\sec^2\theta}\,\mathrm d\theta$$ I tried $u=\tan\theta$ and got $$\int_0^1\frac{\operatorname{arccot}\sqrt{1-u^2}}{u^2+1}\mathrm du$$ Then, I tried integration by parts but a complicated integral came. I think, this is an Ahmed-type integral, because in the problem they occured already two times. I don't have much experience with these type of integrals. WolframAlpha could not find its anti-derivative.

Any hints or suggestions are welcome.

Answer 1

You can solve this by substituting $u=\frac{x}{\sqrt{x^2+1}}$ so that $\frac{du}{1+u^2}=\frac{dx}{(1+2x^2)\sqrt{1+x^2}}$ and $\sqrt{1-u^2}=\frac{1}{\sqrt{1+x^2}}$: $$I=\int_0^1 \frac{\cot^{-1} \sqrt{1-u^2}}{1+u^2}\, du$$ $$=\int_0^{\infty}\frac{\tan^{-1}\sqrt{1+x^2}}{\sqrt{1+x^2}(1+2x^2)}\, dx$$ $$=\int_0^{\infty}\int_0^1 \frac{dt}{(t^2(1+x^2)+1)(1+2x^2)}\, dx$$ $$=\int_0^{\infty}\frac{1}{2+t^2}\int_0^1 \left(\frac{2}{1+2x^2}-\frac{t^2}{x^2t^2+t^2+1}\right)\, dt\, dx$$ $$=2\int_0^{\infty}\frac{dx}{1+2x^2}\int_0^1\frac{dt}{2+t^2}-\int_0^1\frac{t^2}{2+t^2}\int_0^{\infty}\frac{dx}{t^2+1+t^2x^2}\, dt$$ $$=2\frac{\pi}{2\sqrt{2}}\frac{1}{\sqrt{2}}\tan^{-1}\frac{1}{\sqrt{2}}-\frac{\pi}{2}\int_0^1\frac{tdt}{(2+t^2)\sqrt{1+t^2}}$$ $$=\frac{\pi}{2}\tan^{-1}\frac{1}{\sqrt{2}}-\frac{\pi}{2}\int_1^\sqrt{2}\frac{dx}{1+x^2}$$ $$=\pi\tan^{-1}\frac{1}{\sqrt{2}}-\frac{\pi^2}{8}$$

Answer 2

$$\begin{align}\int_0^\frac\pi4\cot^{-1}\sqrt{2-\sec^2x}\,\mathrm dx&=\frac{\pi^2}8-\int_0^\frac\pi4\tan^{-1}\sqrt{2-\sec^2x}\,\mathrm dx\\&=\frac{\pi^2}8-\int_0^1\int_0^\frac\pi4\frac{\sqrt{2-\sec^2x}}{y^2(2-\sec^2x)+1}\mathrm dx\,\mathrm dy\\&=\frac{\pi^2}8-\int_0^1\frac2{2y^2+1}\int_0^\frac1{\sqrt2}\frac1{\sqrt{1-2x^2}}-\frac1{2\left((y^2+1)-(2y^2+1)x^2\right)\sqrt{1-2x^2}}\mathrm dx\,\mathrm dy&\sin x\to x\\&=\frac{\pi^2}8-\int_0^1\frac\pi{\sqrt2(2y^2+1)}-\frac1{(2y^2+1)(y^2+1)}\int_0^\infty\frac{\mathrm dx\,\mathrm dy}{x^2+\frac1{y^2+1}}&\frac{\sqrt{1-2x^2}}x\to x\\&=\frac{\pi^2}8-\frac\pi2\tan^{-1}\sqrt2+\frac\pi2\int_1^\infty\frac{\mathrm dy}{y^2+1}&\frac{\sqrt{y^2+1}}y\to y\\&=\frac{3\pi^2}8-\pi\tan^{-1}\sqrt2\end{align}$$

Answer 3

Putting $u=\sin \theta$ in OP’s integral, we get $$I= \int_0^1\frac{\operatorname{arccot}\sqrt{1-u^2}}{u^2+1}\mathrm du = \int_0^{\frac{\pi}{2}} \frac{\tan ^{-1}\left(\frac{1}{\cos \theta}\right)}{1+\sin ^2 \theta} \cdot \cos \theta d \theta =I(1)$$ where $I(a)= \int_0^{\frac{\pi}{2}} \frac{\tan ^{-1}\left(\frac{a}{\cos \theta}\right)}{1+\sin ^2 \theta} \cdot \cos \theta d \theta$, with derivative w.r.t. $a$ is $$ I^{\prime}(a)= \int_0^{\frac{\pi}{2}} \frac{\cos ^2 \theta}{\left(1+\sin ^2 \theta\right)\left(a^2+\cos ^2 \theta\right)} d \theta= \frac{1}{a^2+2} \int_0^{\frac{\pi}{2}}\left(\frac{2}{2-\cos ^2 \theta}-\frac{a^2}{a^2+\cos ^2 \theta}\right) d \theta $$

Putting $t=\tan \theta$ yields $$ \begin{aligned} I^\prime(a) & =\frac{1}{a^2+2} \int_0^{\infty}\left(\frac{2}{2 t^2+1}-\frac{a^2}{1+a^2+a^2 t^2}\right) d t \\ & =\frac{1}{a^2+2}\left[\sqrt{2} \tan ^{-1}(\sqrt{2} t)-\frac{a}{\sqrt{1+a^2}} \tan ^{-1} \frac{a t}{\sqrt{1+a^2}}\right]_0^{\infty} \\ & =\frac{\pi}{2\left(a^2+2\right)}\left(\sqrt{2}-\frac{a}{\sqrt{1+a^2}}\right) \end{aligned} $$ Integrating $I^\prime(a)$ from $a=0$ to $1$ gives $$ \begin{aligned} I & =\int_0^1 I^{\prime}(a) da\\ & =\frac{\pi}{2}\left[\tan ^{-1}\left(\frac{a}{\sqrt{2}}\right)-\tan ^{-1} \sqrt{1+a^2}\right]_0^1 \\ & =\frac{3 \pi^2}{8}-\pi \tan ^{-1} \sqrt{2} \end{aligned} $$