A goodnight integral problem

[math]\frac{\pi}{4}[/math]

Map x to pi/2 - x
Add to original integral

Welcome to America, the land of the scientists

tan(x) = u^2
sec^2(x)dx = 2*udu
dx = 2*u/(1 + u^4) du = 2*u/[(u^2 - sqrt(2)*u + 1)*(u^2 + sqrt(2)*u +1)]
so the integral becomes integral 0 to infinity 2*u/[(1+u^2)*(u^2 - sqrt(2) + 1)*(u^2 + sqrt(2)*u + 1)] du.
Then, you just do partial fractions here.

I think I saw this in that Inside Interesting Integrals book. Don't you reverse the direction of integration?

Multiply top and bottom by rad(cotx) and then it becomes brainlet tier

>he doesn't write cot
what the fuck do you write? cos/sin? fucking brainlets

cotg

im 26 and i just finished high school so bear with me but doesnt the whole thing simplify to 1?

no
you seem to be doing algebra without a clear understanding of the link between the steps, and instead just guesing what the next line should show

i dont understand. what part did i get wrong?

mathematics isn't a guessing game mate, if you can't tell exactly how you got from one line to another it's going to be wrong.

Since he's being an asshole, the issue is going form the right on the second line after you wrote simplifying to the left on the third line. You missed the +1 with the tanx.

oh i see. thanks

he also has a mistake on the third line when he uses the resiprocal of one of the fractions

why bother fixing the mistakes of someone who is still not at a level where he can follow the principles of mathematical logic? correcting a couple of errors isn't going to make him a better mathematician when he doesn't understand what "therefore" means

theres only one thing more retarded i guess, and that would be posting about it

posting about retards making mistakes makes me retarded? maybe, but at least I can do basic algebra

>resiprocal
reciprocal

гoвopить co мнoй, кoгдa ты мoжeшь гoвopить нa pyccкoм, дypaк блять

You're just being a hostile asshole for no reason. People makes mistakes then they learn from them. You're no different you passive aggressive faggot.

calm down perelman

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I love baiting you asspies lel

Let [math]A \,\in\, \left[0,\, \frac\pi2 \right[[/math]. With [math]t \,=\, \sqrt{\tan\,x}[/math] and a PFD,
[eqn]\int_0^A \frac{\sqrt{\cot\,x}} {\sqrt{\cot\,x} \,+\, \sqrt{\tan\,x}} \,\mathrm dx \,=\, \frac12\, \int_0^\sqrt{\tan\,A} \frac t{t^2 \,+\, 1} \,\mathrm dt \,-\, \frac1{4\,\sqrt2}\, \int_0^\sqrt{\tan\,A} \frac{\sqrt2\,t \,-\, 2}{t^2 \,-\, \sqrt2\,t \,+\, 1} \,\mathrm dt \,-\, \frac1{4\,\sqrt2}\, \int_0^\sqrt{\tan\,A} \frac{\sqrt2\,t \,-\, 2}{t^2 \,+\, \sqrt2\,t \,+\, 1} \,\mathrm dt[/eqn]
The first integral gives:
[eqn]\int_0^\sqrt{\tan\,A} \frac t{t^2 \,+\, 1} \,\mathrm dt \,=\, \frac{\ln \left| \tan\,A \,+\, 1 \right|}2[/eqn]
The second integral gives:
[eqn]\int_0^\sqrt{\tan\,A} \frac{\sqrt2\,t \,-\, 2}{t^2 \,-\, \sqrt2\,t \,+\, 1} \,\mathrm dt \,=\, \frac{\sqrt2}2\, \ln\left| \tan\,A\,\left( 1 \,-\, \sqrt{\frac2{\tan\,A}} \,+\, \frac1{\tan\,A} \right) \right| \,-\, \sqrt2\, \arctan\left( \sqrt{2\,\tan\,A} \,-\, 1 \right) \,-\, \frac{\pi\,\sqrt2}4[/eqn]
Similarly, the third integral gives:
[eqn]\int_0^\sqrt{\tan\,A} \frac{\sqrt2\,t \,-\, 2}{t^2 \,+\, \sqrt2\,t \,+\, 1} \,\mathrm dt \,=\, \frac{\sqrt2}2\, \ln\left| \tan\,A\,\left( 1 \,+\, \sqrt{\frac2{\tan\,A}} \,+\, \frac1{\tan\,A} \right) \right| \,-\, 3\,\sqrt2\, \arctan\left( \sqrt{2\,\tan\,A} \,+\, 1 \right) \,+\, \frac{3\,\pi\,\sqrt2}4[/eqn]
Summing makes the logs converge towards 0, hence
[eqn]\int_0^A \frac{\sqrt{\cot\,x}} {\sqrt{\cot\,x} \,+\, \sqrt{\tan\,x}} \,\mathrm dx \,\underset{A \,\to\, \infty}{\sim}\, \sqrt2\, \arctan\left( \sqrt{2\,\tan\,A} \,-\, 1 \right) \,+\, \frac{\pi\,\sqrt2}4 \,+\, \frac34\, \arctan\left( \sqrt{2\,\tan\,A} \,+\, 1 \right) \,-\, \frac{3\,\pi}{16} \,\xrightarrow[A \,\to\, \infty]{}\, \frac{\pi\,\sqrt2}2 \,+\, \frac{\pi\,\sqrt2}4 \,+\, \frac{3\,\pi}8 \,-\, \frac{3\,\pi}{16}[/eqn]
Therefore,
[eqn]\int_0^{\frac\pi2} \frac{\sqrt{\cot\,x}} {\sqrt{\cot\,x} \,+\, \sqrt{\tan\,x}} \,\mathrm dx \,=\, \frac{12\,\sqrt2 \,+\, 3}{16} \,\pi[/eqn]

"No."

The answer is [math]\frac{\pi}{4}[/math]

Then be useful and find where the mistake is.

see

You need to clear the discontinuity. So multiply the original equation by [math]tan(x) sec^2(x)[/math] and try again.

Lmao
autism doesn't mean you're right

* in the numerator and denominator in case that isn't clear.

>piggot autist is bad at math
Who would've thunk

What's wrong with cot?

test

Hows undergrad treating you

Hey I'm the only one allowed to make these threads.

Trigonometry formulas are complicated enough with 3 functions.

too lazy to download your shit keyboard

Ya ponimayu chto ti pisel, i ya znayu kak govorit po russkiy (no ochen ploho)

You are not smarter than us because you know 2 languages, this is a science board, not linguistics. If you wanna talk in English don't be a dick when some asshole corrects your spelling.

Literally Integration by Parts
sky

What the FUCK do Europeans write then for the reciprocal of tangent?

>Гoвopить

absolutely.

https://www.wolframalpha.com/input/?i=\int^{\pi/2}_{0}+\frac{sqrt(cot(x))}{sqrt(cot(x))+sqrt(tan(x))}+dx

ayyylmao

>not switching to a trivial complex problem and taking the imaginary part of the result
Back to high school.