Integrating Trigonometric Functions with Irrational Exponents

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Homework Statement


The integral from 0 to pi/2 of 1/(1 + (tanx)^sqrt2) dx.


Homework Equations


trig identities?


The Attempt at a Solution


I tried some substitutions but it just made the problem more complicated. I also multiplied by (tanx)^sqrt2 in the numerator and denominator in an attempt to solve it using partial fractions. That ended up with a complex solution which I know is wrong. If it matters I plugged it into my calculator and got about .785. Any nudge in the right direction would be great. Also, is there a way to use equations without having to type it all out?
 
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is series solution an option?
 
We're starting series in a few weeks I think so I'm pretty sure it's not an option. This is for BC Calc and the teacher said that we know how to do it but it will be challenging. My thought is some sort of trig identity but nothing seems to work out.
 
It's a rotten dirty trick question. The integral from 0 to pi/2 of 1/(1+tan(x)^b) is the same as the integral over the same interval of 1/(1+cot(x)^b). Where b is any number. Why? That's the trig identity. Now add them and divide by two.
 
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wow, I knew there was a reason why my teacher smiled when she wrote this on the board. I wrote out the proof of it always equaling pi/4. I also would never have realized that from 0 to pi/2 the tan and cot have the same integral. Thank you so much.
 
yeah that's neat
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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