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What does it mean to 'use the symmetry' in an integration problem?

I was given the following problem: $$ \int_0^{\frac{\pi}{4}} \log(1+\tan \theta) \ d\theta $$

My friend then advised me to 'use the symmetry'

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  • $\begingroup$ Was my question not specific $\endgroup$ Commented Jan 13, 2017 at 19:31
  • $\begingroup$ Your friend was probably suggesting to use the $u$-substitution $u = \frac{\pi}{4} - \theta$. $\endgroup$
    – kobe
    Commented Jan 13, 2017 at 19:32
  • $\begingroup$ Yes, he did something along those lines. But he did this instead $\theta \mapsto \frac{\pi}{4} - \theta $ What is the symmetry thing he talking about??? $\endgroup$ Commented Jan 13, 2017 at 19:34
  • $\begingroup$ When he writes $\theta \mapsto \frac{\pi}{4} - \theta$, he means to replace $\theta$ with $\frac{\pi}{4} - \theta$. More precisely though, you are to use the substitution $u = \frac{\pi}{4} - \theta$. By using this substitution, the limits of integration remain unchanged. $\endgroup$
    – kobe
    Commented Jan 13, 2017 at 19:38
  • $\begingroup$ Yup, I see that. What I'm more curious about is how he spotted that substitution and I believe the clue is in this thing about 'symmetry' and was trying to find out what it meant. @kobe $\endgroup$ Commented Jan 13, 2017 at 19:46

2 Answers 2

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You can use symmetry by substituting $\frac{π}{4}-x$ for x. Then you can use property of logarithms. Everything will cancel out.

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  • $\begingroup$ is that what symmetry is? subbing in $x$ for $\frac{\pi}{4} - x$? $\endgroup$ Commented Jan 13, 2017 at 19:35
  • $\begingroup$ Um, that is (b+a-x), otherwise, yep, that is it. Sometimes we call it the "King" identity, but this might just be an invention of our teacher. It's best to use when this simplifies the integrand, like this one. $\endgroup$
    – Petra
    Commented Jan 14, 2017 at 1:51
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I've always used that for definite integrals we have $\int_a^b f(x)dx=\int_a^b f(b-a-x)dx$. First let your integral equal I and use mentioned fact, $$I=\int_0^{\frac{\pi}{4}} \log(1+\tan \theta) \ d\theta=\int_0^{\frac{\pi}{4}} \log(1+\tan( \frac{\pi}{4}-\theta)) \ d\theta.$$ Now use the tangent subtraction formula to get: $$\int_0^{\frac{\pi}{4}} \ln(1+\frac{\tan(\frac{\pi}{4})-\tan(\theta)}{1+\tan(\frac{\pi}{4}) \tan(\theta)} )d\theta=\int_0^{\frac{\pi}{4}}\ln(\frac{2}{\tan(\theta)+1})d\theta=I.$$ Adding the first and the last expression and using log rules one has, $$2I=\int_0^{\frac{\pi}{4}}\ln(2)d\theta=\frac{\pi \ln(2)}{4},$$ thus $$I= \frac{\pi \ln(2)}{8}.$$

Here's where you can find the tangent subtraction formula I used: http://mathworld.wolfram.com/TrigonometricAdditionFormulas.html

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  • $\begingroup$ Thanks! I never knew about this $\int_a^b f(x) \ dx = \int_a^b f(b-a-x) \ dx $ is there a name for this? Also, when do you find best to use this identity?? $\endgroup$ Commented Jan 13, 2017 at 20:01
  • $\begingroup$ I have never seen a name for that equality because it really is just a special case of a u-substitution, namely letting $x=b-a-u$. It's hard to say when it is "best to use this identity", just try it out on some definite integrals and see where it gets you. After you use it enough you'll start to see when it is useful. I have found it useful with integrands with trig functions and bounds with multiples of $\frac{\pi}{2}$. A fun exercise is to try evaluate the integral $\int_0^{\frac{\pi}{2}} \sin^2(x) dx$ using this method, hint: you will need $\sin^2(x)+\cos^2(x)=1$. $\endgroup$ Commented Jan 14, 2017 at 2:16

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