I need to prove that,

$\displaystyle \tag*{} \int_{0}^{\frac{1}{\sqrt{3}}}\frac{\arctan\left(\frac{1}{\sqrt{1-2x^2}}\right)}{1+x^2} \mathrm{dx}=\frac{13\pi^2}{288}$

Here is what I tried: I tried to solve the integral by Integrating by parts. We have: $\displaystyle \tag{1} \int f(x) g'(x) \mathrm{ dx} = f(x)g(x) - \int f'(x) g(x) \mathrm{ dx}$ I noticed , $\displaystyle \tag*{} \dfrac{1}{1+x^2} = \arctan '(x)$ $\displaystyle \tag*{} g'(x) = \dfrac{1}{1+x^2} \Leftrightarrow g(x) = \arctan(x)$ Now, I defined $f(x)$ $\displaystyle \tag*{} f(x) = \arctan \left( \dfrac{1}{\sqrt{1-2x^2}}\right )$ and $\displaystyle \tag*{} f'(x) = \arctan'\left(\dfrac{1}{\sqrt{1-2x^2}} \right ) = \text{arccot}'(\sqrt{1-2x^2}) = \dfrac{x}{\sqrt{1-2x^2}(1-x^2)}$ Now, using $(1)$ and substituting the values of $f(x)$ and $g(x)$, My indefinite integral becomes: $\displaystyle \tag*{} \arctan(x) \arctan \left (\dfrac{1}{\sqrt{1-2x^2}} \right ) - \int \dfrac{x \arctan(x)}{\sqrt{1-2x^2}(1-x^2)} \mathrm{ dx}$ Now, we want to evaluate: $\displaystyle \tag*{} \int \dfrac{x \arctan(x)}{\sqrt{1-2x^2}(1-x^2)} \mathrm{ dx}$

Now this is integral I am having trouble solving. Any hints or different methods would be greatly appreciated. Thank you.

  • 1
    $\begingroup$ A modification of this answer might be the way to go. $\endgroup$ Feb 3, 2022 at 7:07
  • $\begingroup$ @EeveeTrainer Yes, a bit. Can you help in providing more links which are relevant? $\endgroup$
    – Dhanvin
    Feb 3, 2022 at 15:58

2 Answers 2


A large portion of this answer is taken directly from this answer!

Enforce the substitution

$$y=\sqrt{1-2x^2}\qquad\qquad x^2=\frac {1-y^2}{2}\qquad\qquad\mathrm dx=-\frac {y}{\sqrt{2(1-y^2)}}\,\mathrm dy$$

The integral now becomes

$$\begin{align*}\int\limits_0^{1/\sqrt3}\frac {\arctan\left(\frac 1{\sqrt{1-2x^2}}\right)}{1+x^2}\,\mathrm dx & =\sqrt{2}\int\limits_{1/\sqrt{3}}^1\frac {y\left(\frac {\pi}2-\arctan y\right)}{\sqrt{1-y^2}(3-y^2)}\,\mathrm dy\\ & =\frac {\pi}{\sqrt{2}}\int\limits_{1/\sqrt{3}}^1\frac {y\,\mathrm dy}{\sqrt{1-y^2}(3-y^2)}-\sqrt{2}\int\limits_{1/\sqrt3}^1\frac {y\arctan y}{\sqrt{1-y^2}(3-y^2)}\,\mathrm dy\end{align*}$$

Splitting up the integrand, the first integral can be evaluated with the substitution $y\mapsto\sqrt{1-y^2}$, giving

$$\frac {\pi}{\sqrt{2}}\int\limits_{1/\sqrt{3}}^1\frac {y}{\sqrt{1-y^2}(3-y^2)}\,\mathrm dy=\frac {\pi^2}{12}$$

And enforcing the substitution $t=\sqrt{y}$ on the second integral, we have that

$$\int\limits_0^{1/\sqrt3}\frac {\arctan\left(\frac 1{\sqrt{1-2x^2}}\right)}{1+x^2}\,\mathrm dx=\frac {\pi^2}{12}-\frac 1{\sqrt{2}}\int\limits_{1/3}^1\frac {\arctan\sqrt{t}}{(3-t)\sqrt{1-t}}\,\mathrm dt$$

The last integral is difficult, but using the same methodology and formulas as the answer I have linked above, we first rewrite the integral such that the lower limit is zero.

$$\int\limits_{1/3}^1\frac {\arctan\sqrt t}{(3-t)\sqrt{1-t}}\,\mathrm dt=\underbrace{\int\limits_0^1\frac {\arctan\sqrt{t}}{(3-t)\sqrt{1-t}}\,\mathrm dt}_{J}-\underbrace{\int\limits_0^{1/3}\frac {\arctan\sqrt{t}}{(3-t)\sqrt{1-t}}\,\mathrm dt}_{K}$$

Next, we use the following formula to evaluate the right-hand side

$$\int\limits_0^x\frac {\arctan\sqrt t}{(a-t)\sqrt{b-t}}\,\mathrm dt=\frac 1{\sqrt{a-b}}S\left(\arctan\sqrt{\frac {b-x}{a-b}},\arctan\sqrt{\frac {b+1}{a-b}},\arctan\frac 1{\sqrt{a}}\right)$$

There are two important observations that will help us evaluate the two integrals, namely

  • $S(0,\beta,\gamma)=\pi(\beta-\gamma)$
  • When $\sin^2\alpha+\sin^2\gamma=\sin^2\beta$, then $S(\alpha,\beta,\gamma)=-\alpha^2+\beta^2-\gamma^2$

Substituting $a=3$ and $b=1$, then

$$\begin{align*}J & =\frac 1{\sqrt2}S\left(0,\frac {\pi}4,\frac {\pi}6\right)=\frac {\pi^2}{12\sqrt2}\\K & =\frac 1{\sqrt2}S\left(\frac {\pi}6,\frac {\pi}4,\frac {\pi}6\right)=\frac {\pi^2}{144\sqrt2}\end{align*}$$

Where I have used the first observation to evaluate $J$ and the second observation to evaluate $K$. Taking the difference $J-K$, then

$$\int\limits_{1/3}^1\frac {\arctan x}{(3-x)\sqrt{1-x}}\,\mathrm dx=\frac {11\pi^2}{144\sqrt2}$$

Putting everything together, we get that

$$\int\limits_0^{1/\sqrt{3}}\frac {\arctan\left(\frac 1{\sqrt{1-2x^2}}\right)}{1+x^2}\,\mathrm dx=\frac {\pi^2}{12}-\frac {11\pi^2}{288}\color{blue}{=\frac {13\pi^2}{288}}$$

  • $\begingroup$ Thanks for finishing this integral! I had no hopes. $\endgroup$
    – Dhanvin
    Jun 17, 2022 at 20:22
  • 2
    $\begingroup$ @Dhanvin I think we can both say that pisco is our savior here :P $\endgroup$
    – Frank W
    Jun 17, 2022 at 20:24

Integrate as follows \begin{align} &\int_{0}^{\frac{1}{\sqrt{3}}}\frac{\tan^{-1}\frac{1}{\sqrt{1-2x^2}}}{1+x^2} {dx}\\ \overset{ibp}=&\ \frac{\pi^2}{18}-\int_{0}^{\frac{1}{\sqrt{3}}}\frac{x\tan^{-1} x}{(1-x^2)\sqrt{1-2x^2}}dx \>\>\>\>\>\>\>\sqrt2 x=\cos t\\ =& \ \frac{\pi^2}{18} -\int_{\cot^{-1}\sqrt2}^{\pi/2} \frac{\tan^{-1}(\cos t/\sqrt2)\cos t}{2-\cos^2 t}dt\\ =& \ \frac{\pi^2}{18} -\int_{\cot^{-1}\sqrt2}^{\pi/2}\int_0^1 \frac{\sqrt2\cos^2t}{(2-\cos^2 t)(2+y^2\cos^2t)}dy\ dt\\ =& \ \frac{\pi^2}{18} -\int_0^1\frac1{1+y^2}\bigg(\frac\pi4-\frac\pi{2\sqrt{2+y^2}}+\frac{\cot^{-1} \sqrt{2+y^2}}{\sqrt{2+y^2}}\bigg)dy\\ =& \ \frac{11\pi^2}{144}- \int_0^1 \frac{\cot^{-1} \sqrt{2+y^2}}{\sqrt{2+y^2}}dy =\frac{11\pi^2}{144}-\frac{\pi^2}{32}= \frac{13\pi^2}{288} \end{align} where \begin{align} I=&\int_0^1 \frac{\cot^{-1} \sqrt{2+y^2}}{\sqrt{2+y^2}}dy =\int_0^1 \int_0^1 \frac1{(1+y^2)(y^2+x^2+2)}dx \ dy\\ =& \int_0^1 \int_0^1 \frac1{1+x^2}\bigg(\frac1{1+y^2}-\frac1{y^2+x^2+2}\bigg)dy \ dx =\frac{\pi^2}{16}-I=\frac{\pi^2}{32} \end{align}


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