# Thread: interesting integral problem (from old post)

1. ## interesting integral problem (from old post)

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

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.

2. Originally Posted by oblixps
$\displaystyle \int_0^{\pi/2} \frac{dx}{1+(\tan x)^{\sqrt{2}}}$

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 $\displaystyle \int_0^{\frac{\pi}{2}}\frac{\cos^{\sqrt{2}}(x)}{\c os^{\sqrt{2}}(x)+\sin^{\sqrt{2}}(x)}dx$ and let $\displaystyle z=\frac{\pi}{2}-x$ and add the two integrals together.

3. $\displaystyle \int^{\frac{\pi}{2}}_{0} \frac{dx}{1+\tan^{\sqrt{2}} x} = \int^{\frac{\pi}{2}}_{0} \frac{\cos ^{\sqrt{2}}x \ dx}{\cos ^{\sqrt{2}} x+ \sin^{\sqrt{2}}x}$

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

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

4. Other integrals using the similar technique :

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

and this

$\displaystyle \int_0^{\infty} \frac{dx}{ (x^2 + 1) ( 1 + x^{(\ln(x))^2} )}$

5. Originally Posted by simplependulum
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?

6. Originally Posted by oblixps
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 $\displaystyle x = \pi - u$

7. Originally Posted by Random Variable
let $\displaystyle x = \pi - u$
wow that was so simple i can't believe i didn't think of that. thanks! now to attempt his other integral.

8. would the answer to simplependulum's second integral be pi/2?

9. It should be $\displaystyle \frac{\pi}{4}$.

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

10. $\displaystyle \int_0^{\infty} \frac{dx}{ (x^2 + 1) ( 1 + x^{(\ln(x))^2} )}$
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.

11. or

$\displaystyle \int_0^{\infty} \frac{dx}{ (x^2 + 1) ( 1 + x^{(\ln(x))^2} )} = \int^{\pi/2}_{0} \frac{du}{1+ \tan^{\ln^{2} \tan u}u}$ $\displaystyle = \int^{\frac{\pi}{2}}_{0} \frac{\cos^{\ln^{2} \tan u}u }{\cos ^{\ln^{2} \tan u} u+ \sin^{\ln^{2 \tan u} }u} \ du$

let $\displaystyle x = \frac{\pi}{2} - v$

$\displaystyle = \int^{\frac{\pi}{2}}_{0} \frac{\sin^{\ln^{2} \tan v}v }{\sin^{\ln^{2} \tan v} u+ \cos^{\ln^{2 \tan v} }v} \ dv$

since $\displaystyle \ln^{2} \tan v = (\ln \sin v - \ln \cos v)^{2} = (\ln \cos v - \ln \sin v)^{2} = \ln^{2} \cot v$

so it's exactly like the original problem

12. Originally Posted by Random Variable
or

$\displaystyle \int_0^{\infty} \frac{dx}{ (x^2 + 1) ( 1 + x^{(\ln(x))^2} )} = \int^{\pi/2}_{0} \frac{du}{1+ \tan^{\ln^{2} \tan u}u}$ $\displaystyle = \int^{\frac{\pi}{2}}_{0} \frac{\cos^{\ln^{2} \tan u}u }{\cos ^{\ln^{2} \tan u} u+ \sin^{\ln^{2 \tan u} }u} \ du$

let $\displaystyle x = \frac{\pi}{2} - v$

$\displaystyle = \int^{\frac{\pi}{2}}_{0} \frac{\sin^{\ln^{2} \tan v}v }{\sin^{\ln^{2} \tan v} u+ \cos^{\ln^{2 \tan v} }v} \ dv$

since $\displaystyle \ln^{2} \tan v = (\ln \sin v - \ln \cos v)^{2} = (\ln \cos v - \ln \sin v)^{2} = \ln^{2} \cot v$

so it's exactly like the original problem
oops i was careless, i made those substitutions but forgot the factor of 1/2 outside the integral. thanks!