# interesting integral problem (from old post)

#### 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.

#### Drexel28

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

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.

• oblixps

#### Random Variable

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

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

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

• Also sprach Zarathustra and oblixps

#### simplependulum

MHF Hall of Honor
Other integrals using the similar technique :

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

and this

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

• oblixps

#### oblixps

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?

#### Random Variable

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 $$\displaystyle x = \pi - u$$

#### oblixps

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.

#### oblixps

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

#### Random Variable

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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• oblixps

#### Krizalid

MHF Hall of Honor
$$\displaystyle \int_0^{\infty} \frac{dx}{ (x^2 + 1) ( 1 + x^{(\ln(x))^2} )}$$
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.

• oblixps