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Show that $$\int_{0}^{\sqrt{3}} \sin^{-1} \left( \frac{2x}{1+x^2} \right) ~dx =\frac{\pi}{\sqrt{3}}.$$

When I do the following integral by parts taking $\sin^{-1}()$ as first function and 1 as second, I get an additional log term, $$I=\int_{0}^{\sqrt{3}} \sin ^{-1} \frac{2x}{1+x^2}~.1~ dx = \left[x\sin^{-1}\left(\frac{2x}{1+x^2}\right)-\ln(1+x^2)\right]_0 ^{\sqrt3} = \frac{\pi}{\sqrt3}-\ln4.$$

Please help me.

Pallavi
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3 Answers3

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Notice that $$\frac{d}{dx} \sin ^{-1}\left ( \frac{2x}{1+x^2} \right)= \frac{2}{1+x^2} ~ \frac{1-x^2}{|1-x^2|}.$$ So the function $f(x)=\sin ^{-1} \frac{2x}{1+x^2}$ is non-differentiable at $x=1$ in the domain of integration $(0,\sqrt{3})$. The given definite integral can be done by parts assuming $f(x)$ to be the first and 1 as second function $$I=\int_{0}^{\sqrt{3}} \sin ^{-1} \frac{2x}{1+x^2}~.1~ dx=\left. x \sin^{-1} \frac{2x}{1+x^2}\right |_{0}^{\sqrt{3}}- \int_{0}^{\sqrt{3}} x~\frac{d}{dx} \sin ^{-1} \left( \frac{2x}{1+x^2} \right)dx. $$ Then due to the said non-differentiability at $x=1$, we break the domin of integration in above as $[0,1]$ and $[1,\sqrt{3}]$: $$I=\left . x \sin^{-1} \frac{2x}{1+x^2}\right |_{0}^{\sqrt{3}}- \int_{0}^{1} \frac{+2x}{1+x^2} dx -\int_{1}^{\sqrt{3}} \frac{-2x}{1+x^2} dx= \frac{\pi}{\sqrt{3}} -\ln 2 + \ln 2= \frac{\pi}{\sqrt{3}}.$$

Z Ahmed
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Another approach let $x=\tan(y)$ then $dx=\sec^2(y)dy$ . So the integral becomes $$\int_0^{\frac{\pi}{3}} \arcsin(\sin(2y))\sec^2y\,dy$$. Note that $\sin(2y)=\dfrac{2\tan(y)}{1+\tan^2(y)}$. Now we don't immediately write $\arcsin(\sin2y)=2y$ we see that after $y=\frac{\pi}{4}$ $2y>\frac{\pi}{2}$ but arcsin is defined only from $[-\frac{\pi}{2},\frac{\pi}{2}]$. So after $\pi/4$ we have $\arcsin(\sin(2y))=\pi-2y$ thus our integral becomes $$\int_0^{\frac{\pi}{4}}2y\sec^2y\,dy+\int_{\frac{\pi}{4}}^{\frac{\pi}{6}}(\pi-2y)\sec^2y\,dy$$ which can be easily found out.

cqfd
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Archis Welankar
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Let $x=\tan\theta$, where $\theta\in[0\frac\pi3]$. Then $\dfrac{2x}{1+x^2}=\dfrac{2\tan\theta}{\sec^2\theta}=2\sin\theta\cos\theta=\sin2\theta$.

\begin{align*} \int_0^\sqrt3\sin^{-1}\left(\frac{2x}{1+x^2}\right)dx&=\int_0^{\frac\pi4}\sin^{-1}(\sin2\theta)\frac{d\tan\theta}{d\theta} d\theta+\int_{\frac\pi4}^{\frac\pi3}\sin^{-1}(\sin2\theta)\frac{d\tan\theta}{d\theta} d\theta\\ &=\int_0^{\frac\pi4}2\theta\frac{d\tan\theta}{d\theta} d\theta+\int_{\frac\pi4}^{\frac\pi3}(\pi-2\theta)\frac{d\tan\theta}{d\theta} d\theta\\ &=\left[2\theta\tan\theta\right]_0^{\frac\pi4}-2\int_0^{\frac\pi4}\tan\theta d\theta+\left[(\pi-2\theta)\tan\theta\right]_{\frac\pi4}^{\frac\pi3}+2\int_{\frac\pi4}^{\frac\pi3}\tan\theta d\theta\\ &=\frac\pi2-2\left[\ln\sec\theta\right]_0^\frac\pi4+\frac{\sqrt3\pi}{3}-\frac\pi2+2\left[\ln\sec\theta\right]_{\frac\pi4}^{\frac\pi3}\\ &=\frac{\sqrt3\pi}3-\ln2+\ln2\\ &=\frac{\sqrt3\pi}3 \end{align*}

CY Aries
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