I want to give you this question as an enigma.
Can you prove that the following integral is equal to $\pi$ ?
$$\int_0^\infty \sqrt{\frac{256x^4}{x^{12}+6x^{10}+15x^8+35x^4+6x^2+1}}=\pi.$$
This can be proven without the use of complex numbers.
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Related: http://math.stackexchange.com/questions/266181/a-nasty-integral-of-a-rational-function – Jack D'Aurizio Jun 11 '16 at 17:04
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Thanks, but i want to see if someone can find the solution without the use of complex numbers. – E. Joseph Jun 11 '16 at 17:06
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*Hahahahaha*... – Dementor Jun 11 '16 at 17:06
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4Are you sure it equals $\pi$? – mickep Jun 11 '16 at 17:07
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I agree with @mickep. Mathematica gave me $3.085050028\cdots$ – Ian Miller Jun 11 '16 at 17:11
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5I'm voting to close this question as off-topic because it is an attempt to pose a challenge that the poster already knows (or claims to know) the answer to; thus Not A Real Question. – hmakholm left over Monica Jun 11 '16 at 17:16
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2I don't find this very stimulating, probably there is only a lot of calculations to do (the root will in some way simplified and bla bla bla you express it as the sum of some arctangents and then you have $\pi$) even a lot more if you don't want (why?) to use complex numbers. This is the kind of work that I let the computer do for me. – AlienRem Jun 11 '16 at 17:17
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My guess is that the denominator is supposed to be $$x^{12} + 6x^{10} + 15x^8 + 20x^6 + 15x^4 + 6x^2 + 1 = (x^2 + 1)^6$$ and the coefficient $20$ was somehow added to the wrong place. In that case, your integral is just $$\int_0^{\infty} \frac{16x^2}{(x^2 + 1)^3} \, \mathrm{d}x,$$ which is easy to solve by substituting $x = \tan(\theta)$: $$\int_0^{\pi/2} \frac{16 \tan(\theta)^2 \sec(\theta)^2}{\sec(\theta)^6} \, \mathrm{d}\theta = 16 \int_0^{\pi/2} \sin(\theta)^2 \cos(\theta)^2 \, \mathrm{d}\theta = \pi.$$ As mentioned in the comments, the given integral is incorrect.
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