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$\int \sin^{-1}\dfrac{2x}{1+x^2}dx$

My attempt is as follows:-

$$x=\tan\theta$$ $$dx=\sec^2\theta d\theta$$

$$\int \sin^{-1}(\sin2\theta) \cdot\sec^2\theta d\theta$$

So here should we make cases on the basis of values of $\theta$ or can we write $\sin^{-1}(\sin2\theta)$ as $2\theta$?

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The fact that $\arcsin(\sin(2\theta))=2\theta$ holds or not depends on the range of $\theta$, hence on the range of $x$.
Actually $\arcsin(\sin z)=z$ holds for $z\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$, so we get $\theta\in\left[-\frac{\pi}{4},\frac{\pi}{4}\right]$ and if the integration range is a subrange of $[-1,1]$ we are allowed to state $$ \int \arcsin\left(\frac{2x}{1+x^2}\right)\,dx \stackrel{x\mapsto \tan\theta}{=}\int \frac{2\theta}{\cos^2\theta}\,d\theta\stackrel{\text{IBP}}{=}C+2\left[\log\cos\theta+\theta\tan\theta\right] $$ and $$2\left[\log\cos\theta+\theta\tan\theta\right]=2\left[\log\cos\arctan(x)+x\arctan(x)\right]=2x\arctan(x)-\log(1+x^2). $$

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  • $\begingroup$ but here we don't know what is $x$ right? $\endgroup$ – user3290550 Jan 2 at 17:20
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I think it may be easier to apply integration by parts on the original integral (i.e. differentiate the integrand and integrate $1$) giving \begin{align} \int\arcsin{\left(\frac{2x}{1+x^2}\right)}\mathrm{d}x &=x\cdot\arcsin{\left(\frac{2x}{1+x^2}\right)}-\int x\cdot\frac{2\,\text{sign}(1-x^2)}{1+x^2}\mathrm{d}x\\ &=x\cdot\arcsin{\left(\frac{2x}{1+x^2}\right)}+\text{sign}(x^2-1)\cdot\int\frac{2x}{1+x^2}\mathrm{d}x\\ &=x\cdot\arcsin{\left(\frac{2x}{1+x^2}\right)}+\text{sign}(x^2-1)\cdot\ln{(1+x^2)}+C\\ \end{align} It turns out that if we choose $C=-\ln{(2)}\cdot\text{sign}{(x^2-1)}+C'$ then we get a continuous antiderivative namely $$\int\arcsin{\left(\frac{2x}{1+x^2}\right)}\mathrm{d}x=x\cdot\arcsin{\left(\frac{2x}{1+x^2}\right)}+\text{sign}(x^2-1)\cdot(\ln{(1+x^2)}-\ln{(2)})+C'$$

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In integration, when you apply reverse substitution, which is $x = g(t)$, (for example, in your case you substituted $x = \tan(\theta)$), you have to keep in mind that the function you are using in place for $x$ is one-one and onto (and, if not, you have to restrict the range of the input variable i.e. $t$).

So, to cure your problem, what you do is restrict $\theta$ in the interval $[-\frac{\pi}{2}, \frac{\pi}{2}] \implies x \space \in \space \mathbb R.$ Then, you can easily handle the 3 separate cases of $\theta \in [-\frac{\pi}{2}, -\frac{\pi}{4}), \in [-\frac{\pi}{4}, \frac{\pi}{4}], \in (\frac{\pi}{4}, \frac{\pi}{2}]$. The only thing that seperates the cases is what comes in place of $\sin^{-1}\sin(2\theta)$. For example, in the third case, it will be $\pi - 2\theta$. Then, using IBP will solve the problem.

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  • $\begingroup$ Why the downvote? Do tell me if you find anything wrong/confusing. P.S. @Iabbhattacharjee and I have essentially the same solution and I posted it 5 mins before him, so, it doesn’t really make sense to downvote this solution. $\endgroup$ – Hardik Kalra Jan 2 at 19:08
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The function to find an antiderivative of can be rewritten as $$ f(x)=\begin{cases} -\pi-2\arctan x & x<-1 \\[6px] 2\arctan x & -1\le x\le 1 \\[6px] \pi-2\arctan x & x>1 \end{cases} $$ It's sufficient to see that the derivative is $$ f'(x)=\dfrac{1-x^2}{|1-x^2|}\dfrac{2}{1+x^2} $$ Thus an antiderivative is $$ F(x)=\int_0^x f(t)\,dt $$ Since the function $f$ is odd, we can state that $F$ is even, so we can assume $x>0$.

Recalling that, with integration by parts, $$ \int 2\arctan x\,dx=2x\arctan x-\log(1+x^2) $$ we have, for $0\le x\le 1$, $$ F(x)=\Bigl[2t\arctan t-\log(1+t^2)\Bigr]_0^x=2x\arctan x-\log(1+x^2) $$ and $F(1)=\pi/2-\log2$. For $x>1$, we have \begin{align} F(x) &=F(1)+\int_1^x (\pi-2\arctan t)\,dt \\[6px] &=F(1)+\Bigl[\pi t-2t\arctan t+\log(1+t^2)\Bigr]_1^x \\[6px] &=\frac{\pi}{2}-\log2+\pi x-2x\arctan x+\log(1+x^2)-\pi+\frac{\pi}{2}-\log2 \\[6px] &=-2\log2+\pi x-2x\arctan x+\log(1+x^2) \end{align} Thus the most general antiderivative is $F(x)+c$, where $$ F(x)=\begin{cases} 2x\arctan x-\log(1+x^2) & |x|\le 1 \\[6px] -2\log2+\pi|x|-2x\arctan x+\log(1+x^2) & |x|>1 \end{cases} $$

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If $y=\arctan x,x=\tan y$

$f(x)=\sin^{-1}\dfrac{2x}{1+x^2}=\arcsin(\sin2y)$

Using https://en.wikipedia.org/wiki/Inverse_trigonometric_functions#Principal_values

$-\pi<2y<\pi$

So, $f(x)=2y$ if $-\dfrac\pi2\le2y\le\dfrac\pi2$

$f(x)=\pi-2y$ if $-\dfrac\pi2\le2y-\pi\le\dfrac\pi2\iff 2y\ge\dfrac\pi2$

If $f(x)=-\pi-2y$ if $-\dfrac\pi2\le2y+\pi\le\dfrac\pi2\iff 2y\le-\dfrac\pi2$

Now use https://en.wikipedia.org/wiki/Integration_by_parts

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