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Calculate

$$ \int_{0}^{\frac{\pi}{4}}\tan^{-1}\sqrt{\frac{\cos 2x }{2 \cos^2 x}}dx$$

$\bf{My\; Try::}$ Let $\displaystyle I = \int_{0}^{\frac{\pi}{4}}\tan^{-1}\sqrt{\frac{\cos 2x }{2\cos^2 x}}dx = \int_{0}^{\frac{\pi}{4}}\frac{\pi}{2}-\int_{0}^{\frac{\pi}{4}}\cot^{-1}\sqrt{\frac{\cos 2x}{2\cos^2 x}}dx$

Using The formula $\displaystyle \tan^{-1}(x)+\cot^{-1}(x) = \frac{\pi}{2}\Rightarrow \tan^{-1}(x) = \frac{\pi}{2}-\cot^{-1}(x).$

Now Let $\displaystyle J = \int_{0}^{\frac{\pi}{4}}\cot^{-1}\sqrt{\frac{\cos 2x}{2\cos^2 x}}dx = \int_{0}^{\frac{\pi}{4}}\cot^{-1}\sqrt{\frac{\cos^2 x-\sin^2 x}{2\cos^2 x}}dx = \int_{0}^{\frac{\pi}{4}}\cot^{-1}\sqrt{\frac{1}{2}-\frac{\tan^2 x}{2}}dx$

Now How can I solve after that? Help required.

Thanks

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  • $\begingroup$ Typo in last step. tan^2(x)/2 $\endgroup$ – evil999man Apr 11 '14 at 11:33
  • $\begingroup$ This doesn't help, but the numerical value is roughly 0.41123351671205660911810379166150629730473746630197, and, according to oldweb.cecm.sfu.ca/cgi-bin/isc/… this doesn't appear to be a "well known" number. $\endgroup$ – barrycarter Oct 8 '14 at 21:16
  • $\begingroup$ @barrycarter If you search using the integer relation algorithms option instead, it gives the result that the integral $K$ satisfies $0=-24\,K+\pi^2$, which is equivalent to the result found by Tolaso below. oldweb.cecm.sfu.ca/cgi-bin/isc/… $\endgroup$ – David H Oct 19 '14 at 23:57
  • $\begingroup$ Here is an evaluation of the integral by sos440: sos440.tistory.com/212 $\endgroup$ – Pranav Arora Oct 25 '14 at 22:27
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Proposition :

\begin{equation} \int_0^{\Large\frac{\pi}{4}}\arctan\sqrt{\frac{\mu\cos2x}{\cos^2x}}\,dx=\frac{\pi}{2}\left[\arctan\sqrt{2\mu}-\arctan\sqrt{\frac{\mu}{\mu+1}}\right]\quad,\quad\text{for }\,\mu\ge0 \end{equation}

Proof :

Let \begin{equation} I(\mu)=\int_0^{\Large\frac{\pi}{4}}\arctan\sqrt{\frac{\mu\cos2x}{\cos^2x}}\,dx \end{equation} then \begin{align} I'(\mu)&=\partial_\mu\int_0^{\Large\frac{\pi}{4}}\arctan\sqrt{\frac{\mu\cos2x}{\cos^2x}}\,dx\\ &=\frac{1}{2}\int_0^{\Large\frac{\pi}{4}}\frac{\sqrt{\frac{\mu\cos2x}{\cos^2x}}}{\mu+\frac{\cos2x}{\cos^2x}\mu^2}\,dx\\ &=\frac{1}{2}\int_0^{\Large\frac{\pi}{4}}\frac{\sqrt{\mu(1-2\sin^2x)}}{\mu(1-\sin^2x)+(1-2\sin^2x)\mu^2}\cdot\cos x\,dx\\ &=\frac{1}{2\sqrt{2\mu}}\int_0^{\Large\frac{\pi}{2}}\frac{\sqrt{1-\sin^2\theta}}{\left(1-\frac{1}{2}\sin^2\theta\right)+(1-\sin^2\theta)\mu}\cdot\cos \theta\,d\theta\quad\Rightarrow\quad\sin^2\theta=2\sin^2x\\ &=\frac{1}{\sqrt{2\mu}}\int_0^{\Large\frac{\pi}{2}}\frac{\cos^2\theta}{\sin^2\theta+2(1+\mu)\cos^2\theta}\,d\theta\\ &=\frac{1}{\sqrt{2\mu}}\int_0^{\Large\frac{\pi}{2}}\frac{1}{\tan^2\theta+2(1+\mu)}\,d\theta\\ &=\frac{1}{\sqrt{2\mu}}\int_0^{\infty}\frac{1}{t^2+2(1+\mu)}\cdot\frac{1}{t^2+1}\,dt\quad\Rightarrow\quad t=\tan\theta\\ &=\frac{1}{\sqrt{2\mu}(1+2\mu)}\int_0^{\infty}\left[\frac{1}{t^2+1}-\frac{1}{t^2+2(1+\mu)}\right]\,dt\\ &=\frac{1}{\sqrt{2\mu}(1+2\mu)}\left[\frac{\pi}{2}-\frac{\pi}{2\sqrt{2(1+\mu)}}\right]\\ I(\mu)&=\int\frac{1}{\sqrt{2\mu}(1+2\mu)}\left[\frac{\pi}{2}-\frac{\pi}{2\sqrt{2(1+\mu)}}\right]\,d\mu\\ &=\frac{\pi}{2}\int\left[\frac{1}{\sqrt{2\mu}(1+2\mu)}-\frac{1}{\sqrt{2\mu}(1+2\mu)\sqrt{2(1+\mu)}}\right]\,d\mu\\ \end{align} where \begin{align} \int\frac{1}{\sqrt{2\mu}(1+2\mu)}\,d\mu&=\int\frac{1}{1+y^2}\,dy\qquad\Rightarrow\quad y=\sqrt{2\mu}\\ &=\arctan y+C_1\\ &=\arctan\sqrt{2\mu}+C_1 \end{align} and \begin{align} \int\frac{1}{\sqrt{2\mu}(1+2\mu)\sqrt{2(1+\mu)}}\,d\mu&=\int\frac{1}{(1+y^2)\sqrt{2+y^2}}\,dy\qquad\Rightarrow\quad y=\sqrt{2\mu}\\ \end{align} Using \begin{align} \color{blue}{\int \frac{dx}{(x^2+1)\sqrt{x^2+a}}=\frac{1}{\sqrt{a-1}}\tan^{-1}\left(\frac{x\sqrt{a‌​-1}}{\sqrt{x^2+a}}\right)+C} \end{align} It can be derived by using substitution $x=\dfrac{1}{t}$ followed by $z=\sqrt{at^2+1}$. Hence \begin{align} \int\frac{1}{\sqrt{2\mu}(1+2\mu)\sqrt{2(1+\mu)}}\,d\mu&=\arctan\sqrt{\frac{\mu}{\mu+1}}+C_2 \end{align} then \begin{equation} I(\mu)=\frac{\pi}{2}\left[\arctan\sqrt{2\mu}-\arctan\sqrt{\frac{\mu}{\mu+1}}\right]+C \end{equation} For $\mu=0$, we have $I(0)=0$ implying $C=0$. Thus \begin{equation} I(\mu)=\int_0^{\Large\frac{\pi}{4}}\arctan\sqrt{\frac{\mu\cos2x}{\cos^2x}}\,dx=\frac{\pi}{2}\left[\arctan\sqrt{2\mu}-\arctan\sqrt{\frac{\mu}{\mu+1}}\right]\qquad\quad\square \end{equation}


For $\mu=\frac{1}{2}$, we obtain \begin{align} I\left(\frac{1}{2}\right)&=\int_0^{\Large\frac{\pi}{4}}\arctan\sqrt{\frac{\cos2x}{2\cos^2x}}\,dx\\ &=\frac{\pi}{2}\left[\arctan 1-\arctan\left(\frac{1}{\sqrt{3}}\right)\right]\\ &=\frac{\pi}{2}\left[\frac{\pi}{4}-\frac{\pi}{6}\right]\\ &=\frac{\pi^2}{24} \end{align}

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    $\begingroup$ Yuhuu Mr. @O.L. ٩(˘◡˘)۶ $\endgroup$ – Anastasiya-Romanova 秀 Oct 19 '14 at 19:19
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    $\begingroup$ Say what? How on earth did you all these calculations? (+1) for this! $\endgroup$ – Tolaso Oct 19 '14 at 19:41
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    $\begingroup$ @Tolaso: I agree! Nice and interesting solutions from both of you. Twice +1 $\endgroup$ – Markus Scheuer Oct 20 '14 at 7:02
  • $\begingroup$ Mr. @MarkusScheuer & Tolaso Thank you. Mr. Tolaso's solution is also nice although he made a few mistake but it didn't affect the final result $\endgroup$ – Anastasiya-Romanova 秀 Oct 20 '14 at 8:09
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    $\begingroup$ +1 for the unexpected generalization! My gut tells me the $\mu=1$ case might still posess a deceptively simple solution, but you handled the general case very nicely. $\endgroup$ – David H Oct 20 '14 at 20:53
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Answer: $\displaystyle \int_{0}^{\pi/4}\tan^{-1}\sqrt{\frac{\cos 2x}{2\cos^2 x}}\,dx=\frac{\pi^2}{24}$

Proof:

We are making use of $3$ Lemmas which are (quite ) easy to prove:

1. $\displaystyle \int_{0}^{1}\frac{dx}{\sqrt{x^2+2}\left ( x^2+1 \right )}=\frac{\pi}{6} $

Proof: $$\begin{align*} \int_{0}^{1}\frac{dx}{\sqrt{x^2+2}\left ( x^2+1 \right )} &\overset{x=\sqrt{2}\sinh t}{=\! =\! =\! =\! =\! =\! =\!}\int_{0}^{a}\frac{dt}{1+2\sinh^2 t} \\ &= \int_{0}^{a}\frac{dt}{\cosh (2t)}=\int_{0}^{a}\frac{\cosh (2t)}{1+\sinh^2 (2t)}\,dt\\ &=\frac{1}{2}\tanh^{-1}\left ( \sinh (2a) \right ) \\ &=\frac{1}{2}\tanh^{-1}\left ( \sqrt{\left ( 1+2\sinh^2 a \right )^2-1} \right ) \\ &= \frac{1}{2}\tanh^{-1}\sqrt{3}=\frac{\pi}{6}\\ \end{align*}$$

where $\displaystyle a=\sinh^{-1}\frac{1}{\sqrt{2}} $.

2. $\displaystyle \int_{0}^{\infty}\frac{dx}{\left ( x^2+a^2 \right )\left ( x^2+\beta^2 \right )}=\frac{\pi}{2a\beta\left ( a+\beta \right )}$

Proof: $$\begin{align*} \int_{0}^{\infty}\frac{dx}{\left ( x^2+a^2 \right )\left ( x^2+\beta^2 \right )} &=\int_{0}^{\infty}\frac{1}{\beta^2-a^2}\left ( \frac{1}{x^2+a^2}-\frac{1}{x^2+\beta^2} \right )\,dx \\ &= \frac{1}{\beta^2-a^2}\left ( \frac{\pi}{2a}-\frac{\pi}{2\beta} \right )\\ &= \frac{\pi}{2a\beta\left ( a+\beta \right )}\\ \end{align*}$$

3. It also holds by definition that: $\displaystyle \tan^{-1}a= \int_{0}^{1}\frac{a}{1+a^2x^2}\,dx$.

And now we are ready to evaluate the integral. Successively we have:

$$\begin{align*} \int_{0}^{\pi/4}\tan^{-1}\sqrt{\frac{\cos 2\theta}{2\cos^2 \theta}}\,d\theta &=\int_{0}^{\pi/4}\int_{0}^{1}\frac{\sqrt{\frac{\cos 2\theta}{2\cos^2 \theta}}}{1+\left ( \frac{\cos 2\theta}{2\cos^2 \theta} \right )x^2}\,dx \,d\theta\\ &= \int_{0}^{1}\int_{0}^{\pi/4}\frac{\sqrt{1-2\sin^2 \theta}}{2-2\sin^2 \theta+\left ( 1-2\sin^2 \theta \right )x^2}\sqrt{2}\cos \theta \,\,d\theta \,dx\\ &=\int_{0}^{1}\int_{0}^{\pi/2}\frac{\sqrt{1-\sin^2 \phi}}{2-\sin^2 \phi+\left ( 1-\sin^2 \phi \right )x^2}\cos \phi \,\,d\phi \,dx \\ &=\int_{0}^{1}\int_{0}^{\pi/2}\frac{\cos^2 \phi}{\sin^2 \phi+\left ( x^2+2 \right )\cos^2 \phi}\,\,d\phi \,dx \\ &=\int_{0}^{1}\int_{0}^{\pi/2}\frac{d\phi \,dx}{\tan^2 \phi+x^2+2}=\int_{0}^{1}\int_{0}^{\infty}\frac{dx\,dx}{\left ( y^2+x^2+2 \right )\left ( y^2+1 \right )} \\ &=\frac{\pi}{2}\int_{0}^{1}\frac{dy}{\left ( 1+\sqrt{2+y^2} \right )\sqrt{2+y^2}} \\ &=\frac{\pi}{2}\left ( \frac{\pi}{4}-\frac{\pi}{6} \right )=\frac{\pi^2}{24} \end{align*}$$

which checks numerically with the answer given above. If I have any typos, because I typed it so quickly please let me know so that I correct them.

T:

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  • $\begingroup$ This is brilliant answer! It's numerically match with Wolfram Alpha. A small typo, it should be $\displaystyle \tan^{-1}a= \int_{0}^{1}\frac{a}{1+a^2x^2}dx$. +1 $\endgroup$ – Venus Oct 19 '14 at 16:46
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    $\begingroup$ Your 1. is wrong: $$\int \frac {dt}{\cosh(2t)}=\mathrm{atan}(\tanh t)+C$$ Thus the first integral has value $$\mathrm{atan} (\tanh (\mathrm{asinh}\frac1{\sqrt{2}}))= \mathrm{atan} (\sqrt{\frac23}\cdot\frac1{\sqrt{2}})$$ $$=\mathrm{atan}\frac{1}{\sqrt 3}=\frac{\pi}6$$ Notice that $\mathrm{atanh} \sqrt{3}$ is not even defined. $\endgroup$ – Jean-Claude Arbaut Oct 20 '14 at 4:15
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    $\begingroup$ @Jean-ClaudeArbaut In general $$\int \frac{dx}{(x^2+1)\sqrt{x^2+a}}=\frac{1}{\sqrt{a-1}}\tan^{-1}\left(\frac{x\sqrt{a-1}}{\sqrt{x^2+a}}\right)+C$$ It can be derived by using substitution $x=\dfrac{1}{t}$ followed by $u=\sqrt{at^2+1}$ $\endgroup$ – Anastasiya-Romanova 秀 Oct 20 '14 at 8:18
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    $\begingroup$ @Tolaso Anyway, you've made a minor typo for the the bound of integration at 3rd & 4th line in $3$. It should be $\dfrac{\pi}{2}$ instead of $\dfrac{\pi}{4}$ \begin{align} &=\int_{0}^{1}\int_{0}^{\pi/2}\frac{\sqrt{1-\sin^2 \phi}}{2-\sin^2 \phi+\left ( 1-\sin^2 \phi \right )x^2}\cos \phi \,\,d\phi \,dx \\ &=\int_{0}^{1}\int_{0}^{\pi/2}\frac{\cos^2 \phi}{\sin^2 \phi+\left ( x^2+2 \right )\cos^2 \phi}\,\,d\phi \,dx \end{align} $\endgroup$ – Anastasiya-Romanova 秀 Oct 20 '14 at 8:21
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    $\begingroup$ I don't want to be rude but it looks to me that you copied sos440's solution completely present here: sos440.tistory.com/212 $\endgroup$ – Pranav Arora Oct 25 '14 at 22:27

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