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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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    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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    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