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Evaluate $$I=\int_{0}^{\frac{\pi}{4}} \tan^{-1}\left(\sqrt{\frac{\cos 2x}{2\cos^2 x}}\right)\:dx$$

I tried with substitution:

$$\frac{\cos 2x}{2\cos^2 x}=\tan^2 y$$ So we have

$$\sec^2 x=2-2\tan^2 y$$ Differentiating we get:

$$2\sec^2 x\tan x\:dx=-4\tan y\sec^2 y\:dy$$ So

$$dx=\frac{-\tan y\sec^2 y\:dy}{(1-\tan^2 y)\sqrt{1-2\tan^2 y}}$$ So we get $$I=-\int_{\tan^{-1}\left(\frac{1}{\sqrt{2}}\right)}^0\:\frac{y \tan y \sec^2 y\:dy}{(1-\tan^2 y)\sqrt{1-2\tan^2 y}}$$

I am stuck here?

Quanto
  • 120,125
Ekaveera Gouribhatla
  • 14,193
  • Transform your integral into a double integral $\displaystyle \arctan(u)=\int_0^1 \frac{u}{1+t^2u^2},dt$ or $\displaystyle \arctan(1/u)=\int_0^1 \frac{u}{t^2+u^2},dt$ – FDP Mar 30 '20 at 15:46
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    result is $\frac{\pi^2}{24}$ (empirically determined) – FDP Mar 30 '20 at 16:05

3 Answers3

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\begin{align}J&=\int_{0}^{\frac{\pi}{4}} \tan^{-1}\left(\sqrt{\frac{\cos 2x}{2\cos^2 x}}\right)\,dx\\ &=\int_{0}^{\frac{\pi}{4}}\left(\int_0^1 \frac{\sqrt{\frac{2\cos^2 x}{\cos(2x)}}}{a^2+\frac{2\cos^2 x}{\cos(2x)}}\,da\right)\,dx\\ &\overset{u=\tan x}=\int_0^1 \int_0^1 \frac{\sqrt{2}\sqrt{1-u^2}}{\left(a^2(1-u^2)+2\right)(1+u^2)}\,da\,du\\ &=\int_0^1 \left[\frac{\arctan\left(\frac{\sqrt{2}u}{\sqrt{1-u^2}}\right)}{a^2+1}-\frac{\arctan\left(\frac{u}{\sqrt{1+\frac{a^2}{2}}\sqrt{1-u^2}}\right)}{(a^2+1)\sqrt{a^2+2}}\right]_{u=0}^{u=1} \,da\\ &=\frac{\pi}{2}\int_0^1 \frac{1}{1+a^2}\,da-\frac{\pi}{2}\int_0^1 \frac{1}{(a^2+1)\sqrt{a^2+2}}\,da\\ &=\frac{\pi^2}{8}-\frac{\pi}{2}\left[\arctan\left(\frac{a}{\sqrt{a^2+2}}\right)\right]_0^1 \\ &=\frac{\pi^2}{8}-\frac{\pi}{2}\arctan\left(\frac{1}{\sqrt{3}}\right)\\ &=\frac{\pi^2}{8}-\frac{\pi}{2}\times \frac{\pi}{6}\\ &=\boxed{\frac{\pi^2}{24}} \end{align}

NB: I have used $\displaystyle \arctan(1/u)=\int_0^1 \frac{u}{t^2+u^2}\,dt$

FDP
  • 15,643
1

Substitute $\tan x =\sin y$ \begin{align} & \int_{0}^{\pi/4 } \tan^{-1} \sqrt{\frac{\cos2x}{2\cos^2x}}\ dx =\int_0^{\pi/2}\frac{\cos y}{1+\sin^2 y}\tan^{-1}\frac{\cos y}{\sqrt2} \ dy\\ &\>\>\>\>\>=\int_0^{\pi/2}\int_0^{\pi/4} \frac{\sqrt2 \cos^2 y}{(1+\sin^2 y)(2\cos^2 x+ \cos^2y\sin^2 x)}dx\ dy\\ &\>\>\>\>\>= \frac\pi2\int_0^{\pi/4} \bigg(1-\frac{\cos x}{\sqrt{1+\cos^2x}}\bigg)dx=\frac\pi2\bigg(\frac\pi{4}-\frac\pi6\bigg)=\frac{\pi^2}{24} \end{align}

Quanto
  • 120,125
0

By means of various manipulations that can be probably be more immediately condensed, I ended up recovering an integral that appears in another answer.

$$\begin{align*} & \int_0^\tfrac\pi4 \arctan \sqrt{\frac{\cos(2x)}{2\cos^2x}} \, dx \\ &= \int_0^\tfrac\pi4 \arctan \sqrt{1 - \frac12\sec^2x} \, dx \\ &= \int_\tfrac\pi4^\tfrac\pi2 \arctan(\cos y) \frac{\cot y}{\sqrt{2\sin^2y-1}} \, dy & \sin y=\frac1{\sqrt2}\sec x \\ &= \int_0^\tfrac\pi4 \arctan(\sin y) \frac{\tan y}{\sqrt{2\cos^2y-1}} \, dy & y\to\frac\pi2-y \\ &= \int_0^1 \arctan \frac z{\sqrt{1+z^2}} \cdot \frac z{\sqrt{1-z^4}} \, dz & z=\tan y \\ &= \frac\pi4 \arctan\frac1{\sqrt2} - \frac12 \int_0^1 \frac{\arcsin z^2}{\sqrt{1+z^2}\left(1+2z^2\right)} \, dz & \text{by parts} \\ &= \frac\pi4 \arctan\frac1{\sqrt2} - \frac14 \int_0^1 \frac{\arcsin z}{\sqrt z \sqrt{1+z} (1+2z)} \, dz & z\to\sqrt z \\ &= \frac\pi4 \arctan\frac1{\sqrt2} - \frac14 \int_0^1 \frac{\arctan \frac z{\sqrt{1-z^2}}}{\sqrt z \sqrt{1+z} (1+2z)} \, dz & (*) \\ &= \frac\pi4 \arctan\frac1{\sqrt2} - \frac1{\sqrt2} \int_0^1 \frac{w \arctan \frac{1-w^2}{2w}}{\sqrt{1-w^2} \left(w^2-3\right)} \, dw & w=\sqrt{\frac{1-z}{1+z}} \\ &= \int_0^1 \frac{\sqrt2\,w \arctan w}{\sqrt{1-w^2} \left(w^2-3\right)} \, dw = \boxed{\frac{\pi^2}{24}} & (**) \end{align*}$$

In $(*)$ we convert $\arcsin$ to its equivalent $\arctan$ form. In $(**)$, just before leaning on the linked result, we apply the identity

$$\arctan \frac{1-x^2}{2x} = \frac\pi2-2\arctan x$$

and find the constant term vanishes, since (e.g. via Euler substitution) we have

$$\int_0^1 \frac{w}{\sqrt{1-w^2}\left(3-w^2\right)}\,dw = \frac1{\sqrt2}\arctan\frac1{\sqrt2}$$

user170231
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