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Evaluate this definite integral $$\int_0^{\frac{\pi}{2}}\frac{\cos(x)}{-\ln(\tan\left(\frac{x}{2}\right))\cos^4\left(\frac{x}{2}\right)}dx$$

I used Weierstrass substitution and then Feynman's integration technique (Differentiation under the integral sign) and then got the answer $2\ln(3)$ and I'm interested to see if there are any faster or alternate methods to evaluate this integral.

Edit: I wrote $\tan\frac{\pi}{2}$ originally instead of $\tan\frac{x}{2}$ and it's now corrected

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  • $\begingroup$ Uh, $\tan\left (\frac{\pi}{2}\right )$ is undefined, so... $\endgroup$ – Mnifldz May 14 at 2:41
  • $\begingroup$ oh sorry I meant tan(x/2) $\endgroup$ – Mathsisfun May 14 at 2:43
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    $\begingroup$ I can see that you're new to this site. You should know that your post was closed, likely because no one is interested in solving this integral for themselves necessarily, but they'd like to help you solve it. Can you share what you've worked on? People aren't likely to respond well to posts that are of the form "here's a problem, solve it for me." To have a successful post you have to share what you've tried. The title is also not helpful (don't use superlatives, just give a brief tagline of the problem at hand). $\endgroup$ – Mnifldz May 14 at 2:48
  • $\begingroup$ Oh ok thanks for your tips! $\endgroup$ – Mathsisfun May 14 at 2:49
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Substitute $t= -\ln(\tan\frac{x}{2})$ to replace the inconvenient log function in the denominator

\begin{align} I= \int_0^{\frac{\pi}{2}}\frac{\cos x}{-\ln(\tan\frac{x}{2})\cos^4\frac{x}{2}}dx =2\int_0^\infty \frac{e^{-t} -e^{-3t}}t dt \end{align}

Then, integrate by parts

\begin{align} I &= 2\ln t(e^{-t} -e^{-3t})|_0^\infty -2\int_0^\infty\ln t(-e^{-t} +3e^{-3t})dt \\ & \overset {u=3t} = 2\int_0^\infty\ln t \> e^{-t}dt - 2\int_0^\infty (\ln u -\ln3) \>e^{-u}du \\ &= 2\ln3 \int_0^\infty e^{-u}du= 2\ln3 \end{align}

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    $\begingroup$ Neat!, It'd be great if you share what suggested that u-substitution so that answer reaches out in a broader sense and not just a mere solution to the given problem. (We see that the integrand has $sec^2$ and so U-subbing $tan$ would help. We also see a $\ln \tan $ term and if you try subbing $t = \ln \tan \frac x2 $ We get the denominator to be Trig ratios in $x$ rather than $\frac x2$ which is very convinient and so this substitution seems very reasonable) $\endgroup$ – RishiNandha_M May 14 at 6:22
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    $\begingroup$ @RishiNandha_M - Just thought that log function in the denominator is inconvenient to deal with, hence the substitution to remove it $\endgroup$ – Quanto May 14 at 6:26
  • $\begingroup$ Thank you so much for the helpful advice! This method is significantly simpler and easier to use and understand, and doesn't take a lot of knowledge in advanced methods of integration! $\endgroup$ – Mathsisfun May 14 at 6:27
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    $\begingroup$ @SoMingChun - glad it helps. you’re welcome $\endgroup$ – Quanto May 14 at 6:28
  • $\begingroup$ @Quanto Would you mind adding that to the Answers' main text? $\endgroup$ – RishiNandha_M May 14 at 6:54
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Hint: \begin{align} I&=\int_{0}^{\frac{\pi}{2}}\frac{\cos^2(\frac{x}{2})-\sin^2(\frac{x}{2})}{-\ln(\tan\left(\frac{x}{2}\right))\cos^4\left(\frac{x}{2}\right)}dx\\ &=2\int_{0}^{\frac{\pi}{2}} \frac{1-\tan^2(\frac{x}{2})}{-\ln(\tan\left(\frac{x}{2}\right))} d(\tan\frac{x}{2})\\ &=2\int_0^1 \frac{1-x^2}{-\ln x} dx \\ &=2\int_{0}^{\infty} \frac{e^{-t}-e^{-3t}}{t} dt \\ &=2\ln 3(Frullani \ integral) \end{align}

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  • $\begingroup$ Wow this one is also very simple and helpful! Thanks! $\endgroup$ – Mathsisfun May 14 at 6:32

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