# evaluating $\int_0^{\sqrt3} \arcsin(\frac{2t}{1+t^2}) \,dt$

\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*}

-
How do you get your first equality? – Arturo Magidin Sep 21 '11 at 19:55
Integration by parts... – Chon Sep 21 '11 at 20:00
$t \arcsin(\frac{2t}{1+t^2})- \int t \frac{2(1-t^2)}{(1+t^2)^2}\frac{1}{\sqrt(1-\frac{4t^2}{(1+t^2)^2})}dt= t\arcsin(\frac{2t}{1+t^2})- \int \frac{2(1-t^2)t}{(1+t^2)\sqrt((t^2-1)^2)}dt=t\arcsin(\frac{2t}{1+t^2})+\int \frac{2t}{1+t^2}dt$ – Chon Sep 21 '11 at 20:33
@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. – Arturo Magidin Sep 21 '11 at 20:36
I'm thinking about whether the 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. – Michael Hardy Sep 21 '11 at 20:42

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.

-
And I confess that took some thinking to spot... – Arturo Magidin Sep 21 '11 at 20:33

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.

-
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... – Arturo Magidin Sep 22 '11 at 4:18
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 $2\pi/3$. – Michael Hardy Sep 22 '11 at 17:10
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 $\ln(x)+C$. – Arturo Magidin Sep 22 '11 at 17:17
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 $x$. – Michael Hardy Sep 22 '11 at 17:21
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. – Arturo Magidin Sep 22 '11 at 17:22

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)$

-
... 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). – Arturo Magidin Sep 21 '11 at 20:41