 3:35 AM
There are so many things to explore in integrals

4 hours later… 7:45 AM
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alright that's enough spam :P
The $\LaTeX$ fractal 8:04 AM
Oh.... my...
XD
@Mathphile care to give me an integral? $$\int_0^{\pi/2} (\tan x)^{1/n}dx$$
:P Ok... I will give this a go
Alright so $u$ substitution
I will ignore the $1/n$ because I don't know how to do that
$$\int \tan x\,\mathrm dx$$ Let $\tan x = \sin x \div \cos x$ and let $u=\cos x$
This is because we can now do $\int \sin x\cdot \frac 1u\,\mathrm dx$
Ok let $f(x)=\frac 1u$ and $\sin x = g^\prime (x)$
Then I believe $g(x)=-\cos x$
so we have $$\int \sin x\cdot \frac 1u\,\mathrm dx = -\frac 1u\cos x -\int f^\prime (x) g(x)\,\mathrm dx$$ I think you are complicating it too much Now since $\frac 1u = u^{-1}$ then I assume $\frac{\mathrm d}{\mathrm du}u^{-1}=-u^{-2}$
@Mathphile actually I think I assigned the wrong function
I should let $g^\prime (x)=\frac 1u$ and $f(x)=\sin x$ instead. $\int \tan x dx =\int \frac{\sin x}{\cos x} dx$
$u=\cos x$ 8:16 AM
If $g^\prime (x)=\frac 1u$ then $g(x)=|\ln u|$ @MrPie yes (I won't plus C because I will incorporate $0$ and $\pi/2$ in the end) you're almost there Ok: let $f(x)=\sin x$ and $g^\prime (x)=\frac 1u$, then $g(x)=|\ln u|$
$$\therefore \int \sin x\cdot \frac 1u\,\mathrm dx =|\ln u|\cdot\sin x + \int \cos x \frac 1 u\,\mathrm dx$$
Aha! Now since $u=\cos x$ the integral is just $x$ nope
wait 8:20 AM
Oh facepalm I forgot one bad thing $u=\cos x$ is right I let $u=\cos x$ but I was doing $\mathrm dx$
I need to do $\mathrm du$ btw why are you trying to use integration by parts? I don't know. I just did $\frac{du}{dx}=-\sin x$ 8:22 AM
Yes $dx=-\frac{du}{\sin x}$ And now since $u=\cos x$ we have $$\int \frac{\sin x}{\cos x}\mathrm dx=-\int\frac 1u\mathrm du$$
Wait, it's negative @MrPie yes AHH $\int \frac{1}{u}du = log (u)$ 8:24 AM
And now we have $$\int = -|\ln u|=|\ln u^{-1}|=|\ln \sec x|$$ yes But that doesn't make sense but you cannot substitute the limits in this case because for $n=1$ the integral is divergent Also $\sec \frac{\pi}{2}$ is undefined
Since $\cos \frac{\pi}{2}=0$ yes
therefore this integral is only defined for $n\gt 1$
$$\int_0^{\pi/2} (\tan x)^{1/n}dx=\frac{\pi}{2 \sin{\frac{(n+1)\pi}{2n}}}$$ 8:28 AM
Is this integral defined at all? @MrPie yes for $n\gt 1$ it is an improper integral I thought $$\int_0^{\pi/2} (\tan x)^{1/n}\,\mathrm dx = \left[(n+1)|\ln \sec x|\right]_0^{\pi/2}$$ @MrPie nope that's wrong I thought so integration does not work like differenciation
you cannot apply something like the chain rule 8:32 AM
I can, though, can't I? Just in reverse
$\int x^n\,\mathrm dx=\frac{x^{n+1}}{n+1}$ @MrPie not always Oh wait, I made a mistake anyways.
@Mathphile if we have something like $\int n^x\,\mathrm dx$ then this is $\frac{n^x}{\ln n}+C$ right? Ex. $\int \sin^2(x) \ne \frac{\sin^3(x)}{3}\cos x$ No, first we have to integrate $\sin x$ alone
Which is $\cos x$ @MrPie yes
@MrPie and then? 8:36 AM
How do we apply the power of $2$ here? Is $\int\sin^2(x)\,\mathrm dx=\cos^2(x)$? @MrPie nope Or $\frac{\cos^3(x)}{3}$
I don't get it, and you are right. It's not like $\int \sin x\,\mathrm dx =\frac{\sin^2(x)}{2}$ :P @MrPie yup yup to which message? Hint: use $\sin^2 x = \frac{1- \cos (2x)}{2}$ 8:39 AM
Half angle... @MrPie your message before this one I could also use integration by parts lol @MrPie you could $$\int sin^2 x\,\mathrm dx = \frac 12\int (1-\cos 2x)\,\mathrm dx=\frac 12\left(x-\frac 12\sin (2x)\right) +C$$
It's indefinite so we must consider the whole class of integrals
@Mathphile that's not what I get with integration by parts, though...
Ok int by parts I get $$\int\sin^2 x\,\mathrm dx=\sin x\cdot \cos x-\int \cos^2 x\,\mathrm dx$$
Then I can sub in the integral on the RHS, $\cos^2x = 1-\sin^2 x$
$$\int\sin^2 x\,\mathrm dx \stackrel{?}{=}\frac 12(x-\sin x\cdot \cos x)$$
@Mathphile I think this is it. 8:57 AM
@MrPie yup

8 hours later… 4:37 PM
@MrPie You forgot to negate the RHS. Otherwise, your expression means that $\int\sin^2+\cos^2=\sin\cdot\cos\implies x+C=\sin x\cos x\forall x$.

4 hours later… 9:06 PM
@TheSimpliFire Hi, I found an interesting question on this site.
For $b=171$ , I did not find a solution yet. — Peter 7 hours ago @Peter That is an interesting one indeed.
@Mathphile See here, I found an even easier proof for your integral, generalising for all $x^m/(x^n+1)$.
0  Thanks to @Lord Shark the Unknown, @Ali Shather, and @Dr Zafar Ahmed DSc, I was able to prove it. Let $x^n=y$, then $dx=\frac{y^{1/n-1}}{n}dy$ , \int_{0}^{\infty}\frac{x^{m}}{x^n+1}dx =\frac{1}{n}\int_{0}^{\infty}\frac{y^{m/n+1/n-1}}{y+1}dy =\frac{1}{n}\mathrm{B}(m/n+1/n,1-(m/n+1/n)) =\frac...

No need to go through all the complex analysis XD 9:36 PM
@TheSimpliFire Do you think we can prove that for 171, there is no solution ?