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Evaluate the integral ratio$$\dfrac{I_1}{I_2}=\dfrac{\displaystyle \int_{0}^{\frac{\pi}{2}} \sqrt{1+\frac{1}{\sqrt{1+\tan^nx}}} \mathrm{d}x}{\displaystyle \int_{0}^{\frac{\pi}{2}} \sqrt{1-\frac{1}{\sqrt{1+\tan^nx}}} \mathrm{d}x}$$

Using Wolframalpha, I found out that the ratio of the above two integrals is $\sqrt{2}+1$, that is, independent of $n$. I couldn't think of a way to solve it.

I don't think any simple substitutions would work here. Any useful hints would be appreciated.

V.G
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  • Did you try taking the derivative with respect to $n$ and trying to show that is 0? – Michael Burr Dec 08 '20 at 05:41
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    Well, I didn't try that...but wouldn't that be a tedious task? Initially, the question was in the form of a particular value assigned to $n$, so one wouldn't even know that it is same for all $n$ if one doesn't take the help from computer. So, I am hoping for a method that would yield the answer by computing it... – V.G Dec 08 '20 at 05:42
  • what is the source of problem? – Hari Ramakrishnan Sudhakar Dec 08 '20 at 06:03
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    @AlbusDumbledore A friend gave it to me. – V.G Dec 08 '20 at 06:23
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    Often the change of variables $x\mapsto \frac\pi2-x$ can be helpful here ($\tan x$ changes to $\cot x$, and then algebra allows the new function to look very similar to the old function). I couldn't make it work though. – Greg Martin Dec 08 '20 at 06:30
  • @GregMartin Ofcourse, that would be the first instinct for everyone who will see this kind of problem, but I also don't see that to be very helpful here. – V.G Dec 08 '20 at 06:31
  • I think there is some trick like first finding $\frac{I_1+I_2}{I_1-I_2}$ and then finding ratio – Hari Ramakrishnan Sudhakar Dec 08 '20 at 07:22
  • Looks, like the solution lies in some clever change of variables, like $\cos^2 \theta = \frac{1}{\sqrt{1 + \tan^n x}}$. The result mentioned can be obtained in the limit $n \rightarrow \infty$. In this case $$\tan^n x = \begin{cases}0, x < \frac{\pi}{4} \ \infty, x > \frac{\pi}{4}\end{cases}$$. So the numerator would give : $\frac{\pi}{4} (\sqrt{2} + 1)$, and the denominator $\frac{\pi}{4} $, which is desired $\sqrt{2} + 1$. However it is only limiting case. – spiridon_the_sun_rotator Dec 08 '20 at 10:01
  • @spiridon_the_sun_rotator, If $x > \frac{\pi}{4}$, then $\tan^n x=\infty$, so wouldn't that imply that ratio will tend to $1$? – V.G Dec 08 '20 at 10:23
  • @Vilakshan on the interval $(0, \pi / 4)$ the integrand of the upper integral is $\sqrt{2}$, and on the interval $(\pi 4, \pi / 2)$ it is $1$. For the lower integral $1$ and $0$ correspondingly. – spiridon_the_sun_rotator Dec 08 '20 at 10:50

1 Answers1

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Note

\begin{align} \frac{I_1}{I_2}&=1+ \frac{I_1-I_2}{I_2} \\ &=1+\frac1{I_2}\int_0^{\frac\pi2} \left( \sqrt{1+\frac{1}{\sqrt{1+\tan^nx}}}- \sqrt{1-\frac{1}{\sqrt{1+\tan^nx}}} \right)dx \\ &=1+\frac1{I_2}\int_0^{\frac\pi2} \frac{\sqrt{\sqrt{\sin^nx+\cos^nx}+\sqrt{\cos^nx}} - \sqrt{\sqrt{\sin^nx+\cos^nx}-\sqrt{\cos^nx}}}{\sqrt[4]{\sin^nx+\cos^nx}}dx \\ &=1+\frac1{I_2} \int_0^{\frac\pi2} \frac{\sqrt{2\sqrt{\sin^nx+\cos^nx}-2\sqrt{\sin^nx}} }{\sqrt[4]{\sin^nx+\cos^nx}}dx \\ &= 1+\frac{\sqrt2}{I_2} \int_0^{\frac\pi2} \sqrt{1-\frac{1}{\sqrt{1+\cot^nx}}}dx\>\>\>\>\>\>\>\>(x\to \frac\pi2-x)\\ &=1+\frac{\sqrt2}{I_2}\int_0^{\frac\pi2} \sqrt{1-\frac{1}{\sqrt{1+\tan^nx}}}dx \\ &=1+\sqrt2 \\ \end{align}

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