# Evaluate the integral $\int_{\large \frac13}^3\frac{\sin^{-1}\frac x{\sqrt{1+x^2}}}xdx$

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?

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

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 $$\int_{1/3}^3\frac{\ln x}{1+x^2}\: dx=-\int_{1/3}^3\frac{\ln u}{1+u^2}\: du=0.$$
• May be, you could precise why the last integral is $0$. Cheers. Commented Sep 2, 2016 at 7:17