interesting integral problem (from old post)

Aug 2008
249
20
\(\displaystyle \int_0^{\pi/2} \frac{dx}{1+(\tan x)^{\sqrt{2}}}\)

This problem was found on an old post and i thought it was an interesting problem. a hint was given which said remember that tan x = cot(pi/2 - x). i tried letting u = pi/2 - x and i got:
\(\displaystyle \int_0^{\pi/2} \frac{du}{1+(\cot u)^{\sqrt{2}}}\)

i'm not sure what to do next though. my substitution didn't seem to do much.
 

Drexel28

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Nov 2009
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\(\displaystyle \int_0^{\pi/2} \frac{dx}{1+(\tan x)^{\sqrt{2}}}\)

This problem was found on an old post and i thought it was an interesting problem. a hint was given which said remember that tan x = cot(pi/2 - x). i tried letting u = pi/2 - x and i got:
\(\displaystyle \int_0^{\pi/2} \frac{du}{1+(\cot u)^{\sqrt{2}}}\)

i'm not sure what to do next though. my substitution didn't seem to do much.
This is famous. Write it as \(\displaystyle \int_0^{\frac{\pi}{2}}\frac{\cos^{\sqrt{2}}(x)}{\cos^{\sqrt{2}}(x)+\sin^{\sqrt{2}}(x)}dx\) and let \(\displaystyle z=\frac{\pi}{2}-x\) and add the two integrals together.
 
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May 2009
959
362
\(\displaystyle \int^{\frac{\pi}{2}}_{0} \frac{dx}{1+\tan^{\sqrt{2}} x} = \int^{\frac{\pi}{2}}_{0} \frac{\cos ^{\sqrt{2}}x \ dx}{\cos ^{\sqrt{2}} x+ \sin^{\sqrt{2}}x} \)

let \(\displaystyle x = \frac{\pi}{2} -u \)

\(\displaystyle = - \int_{\frac{\pi}{2}}^{0} \frac{\cos ^{\sqrt{2}} (\pi/2 -u) \ du}{\cos ^{\sqrt{2}} (\pi /2 -u )+ \sin^{\sqrt{2}} (\pi /2 -u)} \)

\(\displaystyle = \int^{\frac{\pi}{2}}_{0} \frac{\sin ^{\sqrt{2}} u \ du}{\sin ^{\sqrt{2}} u + \cos^{\sqrt{2}} u}\)

so \(\displaystyle \int^{\frac{\pi}{2}}_{0} \frac{dx}{1+\tan^{\sqrt{2}} x} \ dx = \frac{1}{2} \Big( \int^{\frac{\pi}{2}}_{0} \frac{\cos ^{\sqrt{2}}x \ dx}{\cos ^{\sqrt{2}} x+ \sin^{\sqrt{2}}x} + \int^{\frac{\pi}{2}}_{0} \frac{\sin ^{\sqrt{2}} x \ dx}{\sin ^{\sqrt{2}} x + \cos^{\sqrt{2}} x} \Big)\)

\(\displaystyle = \frac{1}{2} \int^{\frac{\pi}{2}}_{0} \ dx = \frac{\pi}{4} \)
 

simplependulum

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Other integrals using the similar technique :

\(\displaystyle \int^{\pi}_0 \frac{dx}{ 1 + (\sin(x))^{\cos(x)}}\)

and this

\(\displaystyle \int_0^{\infty} \frac{dx}{ (x^2 + 1) ( 1 + x^{(\ln(x))^2} )}\)
 
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Aug 2008
249
20
Other integrals using the similar technique :

\(\displaystyle \int^{\pi}_0 \frac{dx}{ 1 + (\sin(x))^{\cos(x)}}\)

and this

\(\displaystyle \int_0^{\infty} \frac{dx}{ (x^2 + 1) ( 1 + x^{(\ln(x))^2} )}\)
i'm having some trouble with the first integral you posted. i tried turning sin(x) into cos(x)/cot(x) and then using the substitution x = pi/2 - u, i then added the two integrals together and divided by 2. but i couldn't simplify the integral like shown above for the first integral i posted. i also tried making some u substitutions first but all they did was make the integral more complicated. how should this integral be tackled?
 
May 2009
959
362
i'm having some trouble with the first integral you posted. i tried turning sin(x) into cos(x)/cot(x) and then using the substitution x = pi/2 - u, i then added the two integrals together and divided by 2. but i couldn't simplify the integral like shown above for the first integral i posted. i also tried making some u substitutions first but all they did was make the integral more complicated. how should this integral be tackled?
let \(\displaystyle x = \pi - u \)
 
Aug 2008
249
20
would the answer to simplependulum's second integral be pi/2?
 
May 2009
959
362
It should be \(\displaystyle \frac{\pi}{4} \).

EDIT: Make the substitution \(\displaystyle x = \tan u \), then the integral becomes \(\displaystyle \int^{\pi/2}_{0} \frac{du}{1+ \tan^{\ln^{2} \tan u}u} \) (Wondering)
 
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Krizalid

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Mar 2007
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Santiago, Chile
\(\displaystyle \int_0^{\infty} \frac{dx}{ (x^2 + 1) ( 1 + x^{(\ln(x))^2} )}\)
put \(\displaystyle x\mapsto\frac1x\) and the integral becomes \(\displaystyle \int_{0}^{\infty }{\frac{x^{\ln ^{2}x}}{\left( x^{2}+1 \right)\left( 1+x^{\ln ^{2}x} \right)}\,dx},\) now sum the original integral and this one to get the result.
 
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