Can't Figure Out This Integral? Let's Help Each Other!

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SUMMARY

The integral discussed is \int_{0}^{\frac{\pi }{2}}\, \frac{1}{1+(tanx)^{\sqrt 2}} dx, which is known for its complexity and has been featured in old Putnam exams. A key technique for solving this integral involves the substitution \frac{\pi}{2} - x into the function f(x) = \frac{1}{1+tan^{\sqrt(2)}(x)}, leading to the identity f(π/2 - x) = 1-f(x). This allows the integral to be split into two parts, simplifying the evaluation process.

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Hello everyone,

I brushing up on integration techniques and I came across this problem in a book. Does anyone here know were to start? Even Wolfram blanked on it!

\int_{0}^{\frac{\pi }{2}}\, \frac{1}{1+(tanx)^{\sqrt 2}} dx

This integral appeared in the book before sequences and series, so the book did not intend to use any type of series expansion.

Let me know!
 
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This is an integral from one of the old putnams, if you google around you can probably find it.

Consider the substitution of \frac{\pi}{2} - x into f(x) = \frac{1}{1+tan^{\sqrt(2)}(x)} This gives f(π/2 - x) = 1-f(x) Now, break the integral up into two parts, one from 0 to pi/4 and the other from pi/4 to pi/2 and manipulate the second integral using the above identity.
 
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Thanks for the reply,

You're right, I found the answer quickly after a google search. Very interesting integral!
 

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