How do I integrate the following function? $$\int_0^\frac{\pi}{4}\ln(\tan^2 x+1) \; dx$$
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Thank you but I think that it must be a solution without Catalan's constant. – Umbrita Andrei Sep 11 '15 at 19:28
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I don't know a way of expressing the Catalan constant in other constants. Why do you think there must be a way? – mickep Sep 11 '15 at 19:36
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This problem is from a book for 12th students and I don't learn anything about this constant at class – Umbrita Andrei Sep 11 '15 at 19:38
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Maybe there should be $\frac{\pi}{2}$ instead of $\frac{\pi}{4}$? Otherwise I don't think you can get away with it without catalan constant – Oussama Boussif Sep 11 '15 at 19:40
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The book says it's pi/4... – Umbrita Andrei Sep 11 '15 at 19:40
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Is there an answer in your book? – mickep Sep 11 '15 at 19:44
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Unfortunately not for this problem:) – Umbrita Andrei Sep 11 '15 at 19:45
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Well, there is not much to do, the answer for this integral is $\frac{\pi}{2}\log 2-G$, where $G$ is the Catalan number. – mickep Sep 11 '15 at 19:50
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$$\int_{0}^{\frac{\pi}{4}}\log(\tan^2(x)+1)\,dx = -2\int_{0}^{\frac{\pi}{4}}\log(\cos\theta)\,d\theta $$ then exploiting the Fourier expansion of $\log(\cos\theta)$ it follows that:
$$\int_{0}^{\frac{\pi}{4}}\log(\tan^2(x)+1)\,dx =\color{red}{-K+\frac{\pi}{2}\,\log 2}$$ where $K$ is Catalan's constant.
Jack D'Aurizio
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