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$$\begin{align*} \int \arcsin\left(\frac{2t}{1+t^2}\right)\,dt&=t\arcsin\left(\frac{2t}{1+t^2}\right)+\int\frac{2t}{1+t^2}\,dt\\ &=t\arcsin\left(\frac{2t}{1+t^2}\right) + \ln(1+t^2)+C \end{align*}$$ So $$ \int\nolimits_0^{\sqrt3} \arcsin\left(\frac{2t}{1+t^2}\right)=\pi/\sqrt3+2\ln2.$$

However the result seems to be $ \pi/\sqrt3 $ only. Why is there this $ 2\ln2 $?

Detail:

$$ \begin{align*} t \arcsin\left(\frac{2t}{1+t^2}\right)&- \int t \left(\frac{2(1-t^2)}{(1+t^2)^2}\right)\frac{1}{\sqrt{1-\frac{4t^2}{(1+t^2)^2}}}\,dt\\ &= t\arcsin\left(\frac{2t}{1+t^2}\right)- \int \frac{2(1-t^2)t}{(1+t^2)\sqrt{(t^2-1)^2}}\,dt\\ &=t\arcsin\left(\frac{2t}{1+t^2}\right)+\int \frac{2t}{1+t^2}\,dt \end{align*} $$

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    How do you get your first equality?2011-09-21
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    Integration by parts...2011-09-21
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    $ t \arcsin(\frac{2t}{1+t^2})- \int t \frac{2(1-t^2)}{(1+t^2)^2}\frac{1}{\sqrt(1-\frac{4t^2}{(1+t^2)^2})}dt= t\arcsin(\frac{2t}{1+t^2})- \int \frac{2(1-t^2)t}{(1+t^2)\sqrt((t^2-1)^2)}dt=t\arcsin(\frac{2t}{1+t^2})+\int \frac{2t}{1+t^2}dt $2011-09-21
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    @PlaneChon-Ju: Yes, you cancelled $1-t^2$ with $\sqrt{(t^2-1)^2}$ as $-1$, but that does not hold over your entire interval.2011-09-21
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    I'm thinking about whether the [Weierstrass substitution](http://en.wikipedia.org/wiki/Weierstrass_substitution) might do something. We have $\arcsin \frac{2t}{1+t^2} = x$ and $t = \tan(x/2) = \frac{\sin x}{1+\cos x}$, so $dt = \frac{1+t^2}{2}\;dx= t\sin x\;dx$, and the integral becomes $\int_0^{\pi/3} \frac{x\;dx}{1+\cos x}$, and it still looks as if an integration by parts is needed. I'm not sure whether it's looking any better.2011-09-21
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    If you split the integral and are careful with the signs, you'll see that the answer is indeed just $\pi/\sqrt{3}$.2011-09-21
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    In fact, putting in the limits will show you that the integration by parts leads to an improper integral with an infinite discontinuity at $1$, which makes sense given that $\arcsin(x)$ has no derivative (not even a one-sided derivative) at $x=1$ (which corresponds to $t=1$). So the integral needs to be evaluated by splitting anyway.2011-09-22

3 Answers 3

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I think you made a simplification error. We have $$\begin{align*} \frac{d}{dt}\arcsin\left(\frac{2t}{1+t^2}\right) &= \frac{1}{\sqrt{1 - \frac{4t^2}{(1+t^2)^2}}}\left(\frac{2t}{1+t^2}\right)'\\ &= \frac{(1+t^2)}{\sqrt{(1+t^2)^2-4t^2}}\left(\frac{2(1+t^2)-4t^2}{(1+t^2)^2}\right)\\ &= \frac{(1+t^2)}{\sqrt{(1-t^2)^2}}\left(\frac{2(1-t^2)}{(1+t^2)^2}\right)\\ &= \frac{2(1-t^2)}{(1+t^2)\sqrt{(t^2-1)^2}} =\frac{2(1-t^2)}{(1+t^2)|t^2-1|}. \end{align*}$$ You then cancelled to get $$-\frac{2}{1+t^2}.$$ However, that cancellation is only valid if $t^2-1\geq 0$, i.e., if $|t|\geq 1$. Yet your integral covers a region that includes places where you get $t^2\lt 1$, so that the cancellation is not valid over the entire interval. Try doing it by splitting the integral as an integral over $[0,1]$ and over $[1,\sqrt{3}]$, being careful with the signs.

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    And I confess that took some thinking to spot...2011-09-21
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The Weierstrass substitution in a slightly different form from that in which I'm accustomed to seeing it will do it.

We have $$ \begin{align} t & = \tan\frac x2 \\ \\ \frac{2\;dt}{1+t^2} & = dx \\ \\ \frac{2t}{1+t^2} & = \sin x \\ \\ \frac{1-t^2}{1+t^2} & = \cos x \end{align} $$ That's the usual Weierstrass substitution. Now, differentiate the first line above to get $$ dt = \frac12\sec^2\frac x2\;dx $$ so this is $$ \frac{dx}{2\cos^2\frac x2} $$ and by the cosine half-angle formula, this is $$ \frac{dx}{1+\cos x}. $$ By the third line above, we have $$ \arcsin\left(\frac{2t}{1+t^2}\right) = \arcsin \sin x = x $$ (if $0\le x\le \pi/2$). Therefore the desired integral becomes $$ \int \frac{x\;dx}{1+\cos x} = \int x\;dt. $$ Integrating by parts, we get $$ xt - \int t\;dx = x\tan\frac x2 - \int \tan \frac x2 \; dx = x\tan\frac x2 - 2\log\cos\frac x2 + C. $$ As $t$ goes from $0$ to $\sqrt{3}$, $x$ goes from $0$ to $\pi/3$, and there you have it.

Correction: As $t$ goes from $0$ to $\sqrt{3}$, the function $t\mapsto2t/(1+t^2)$ goes from $0$ up to $1$ and then starts going down again. It reaches its maximum at $t=1$. So $\sin x$ goes from $0$ up to $1$ and then starts going down again. Thus $x$ goes from $0$ to $2\pi/3$.

This creates problems when one says $\arcsin\sin x = x$, since that applies when $x$ is between $0$ and $\pi/2$. For $x$ between $\pi/2$ and $2\pi/3$, we'd have $\arcsin\sin x = \pi-x$ and we need to examine that interval separately.

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    Isn't $\tan(\pi/6) = \sqrt{3}/3$? If $x=\pi/3$, then $t=\tan(x/2) = \sqrt{3}/3\neq \sqrt{3}$. So I think it should be from $0$ to $2\pi/3$. That means you need to keep the absolute value in the logarithm, so that you get $\log|\cos(x/2)|$, which will force you to be careful in the evaluation. Otherwise, your antiderivative leads to the same issues as the original one, I think...2011-09-22
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    I see: As $t$ goes from $0$ to $\sqrt{3}$, the function $2t/(1+t^2)$ goes from $0$ up to $1$ and then starts going down again. So $\sin x$ goes from $0$ up to $1$ and then starts going down again. Thus $x$ goes from $0$ to $2\pi/3$.2011-09-22
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    And you have to be careful with your integration by parlts, because there is no derivative at $t=1$. You are still missing the absolute value in the logarithm: remember that the integral of $1/x$ is $\ln|x|+C$, not $\ln(x)+C$.2011-09-22
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    So it appear that the thing to do is look at two intervals _separately_: $t$ goes from $0$ to $1$, so that $2t/(1+t^2)$ goes from $0$ to $1$, and then $t$ goes from $1$ to $\sqrt{3}$, and $2t/(1+t^2)$ goes from $1$ down to $\sqrt{3}/2$. And in the latter interval, $\arcsin\sin x= \pi - x$ rather than being equal to $x$.2011-09-22
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    That's one solution. Alternatively, if you work with the definite integral throughout instead of the indefinite integral separately, you'll notice that after integration by parts you end up with an improper integral, which needs to be dealt with carefully.2011-09-22
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    The arcsine can be an unnatural thing, a "multiple-valued function", if you can tolerate that kind of language. _Sometimes_ one wants the particular value of the arcsine to be the right one for the occasion rather than the conventionally defined one.2011-09-22
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    The problem is not the possible multiple values of $\arcsin$; the problem is that $\arcsin$ does not have a derivative (even a one-sided derivative) at $x=1$ or at $x=-1$. So when you do integration by parts with the derivative of the arcsine, you are introducing a "hole", leading to an improper integral. This would be true even if you pick a different branch of the arcsine: you end up with point where it has no derivative.2011-09-22
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    I wasn't suggesting picking a different branch; I was suggesting that a more natural thing to consider (although not the thing denoted in conventional usage by this integral as written) would be to go from one branch to another when the sine stops increasing and begins decreasing. (What particular things one would need to do to evaluate such integral is another question.)2011-09-22
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part of integral solution is $\ln(1+t^2)$. When you insert integral bounds you get $\ln(1+(\sqrt{3})^2)-\ln(1+(0)^2)$$=\ln(4)-ln(1)$$=2\ln(2)$

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    ... which is exactly what the OP said, but which is not the correct answer (the correct answer is in fact just $\pi/\sqrt{2}$; there is no $2\ln(2)$ even thought he antiderivative seems correct).2011-09-21