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I found the following integral on the net. $$\int_{\frac13}^3\frac{\sin^{-1}\frac x{\sqrt{1+x^2}}}x\ dx$$ My approach was putting $x=\tan\theta$, after which the integral reduces to $\theta\cot\theta$. Then what should I do?

I got stuck while applying integration by parts because $\int{\theta\cot\theta}=\theta\int\cot\theta-\int\int\cot\theta$. How should I find the integral of $\int\ln\sin x$ which occurs in the second term?

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  • $\begingroup$ You forgot that $dx = \sec^2{\theta} d\theta$. Thus the integrand is $2 \theta/\sin{(2 \theta)}$. $\endgroup$
    – Ron Gordon
    Commented Sep 2, 2016 at 5:32
  • $\begingroup$ @RonGordon Oh right.Thanks $\endgroup$
    – user220382
    Commented Sep 2, 2016 at 5:35
  • $\begingroup$ @RonGordon But even $2xcosec(2x)$ is'nt easy to integrate ...can you suggest something? $\endgroup$
    – user220382
    Commented Sep 2, 2016 at 5:40
  • $\begingroup$ @OlivierOloa's solution is the way to go. $\endgroup$
    – Ron Gordon
    Commented Sep 2, 2016 at 5:40

1 Answer 1

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Alternatively, one may observe that $$ \left(\arcsin\frac{x}{\sqrt{1+x^2}}\right)'=\frac1{1+x^2}, \quad x \in \mathbb{R}, \tag1 $$ giving $$ \arcsin\frac{x}{\sqrt{1+x^2}}=\arctan x, \quad x \in \mathbb{R}, \tag2 $$ then integrating by parts, one gets $$ \begin{align} I:=\int_{1/3}^3 \frac{\arcsin\dfrac{x}{\sqrt{1+x^2}}}x\:dx &=\left[\ln x\frac{}{} \arctan x\right]_{1/3}^3-\int_{1/3}^3\frac{\ln x}{1+x^2}\: dx \\\\&=\left[\ln x\frac{}{} \arctan x\right]_{1/3}^3-0 \\\\&=\ln 3\cdot \arctan 3+\ln 3\cdot \arctan \frac13 \end{align} $$ then, using $\arctan x+ \arctan \dfrac1x=\dfrac{\pi}2$, $x>0$,

$$ \int_{1/3}^3 \frac{\arcsin\dfrac{x}{\sqrt{1+x^2}}}x\:dx=\frac{\pi}2 \: \ln 3. $$

Remark. By the change of variable $u=\dfrac1x$, $du=-\dfrac{dx}{x^2}$, we have noticed that $$ \int_{1/3}^3\frac{\ln x}{1+x^2}\: dx=-\int_{1/3}^3\frac{\ln u}{1+u^2}\: du=0. $$

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  • $\begingroup$ May be, you could precise why the last integral is $0$. Cheers. $\endgroup$ Commented Sep 2, 2016 at 7:17
  • $\begingroup$ @Claude Leibovici Edited. Thanks! $\endgroup$ Commented Sep 2, 2016 at 13:41

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