Question

Evaluate ${\int }_0^{\sqrt{3}}\frac{1}{1+x^2}.{\sin }^{-1}\left(\frac{2x}{1+x^2}\right)dx$.

• A. $\frac{5}{72}{\pi }^2$
• B. $\frac{13}{144}{\pi }^2$
• C. $\frac{7}{72}{\pi }^2$
• D. $\frac{1}{12}{\pi }^2$

Answer 1

Answer: (C)

$\frac{7}{72}{\pi }^2$

Let $I={\int }_0^{\sqrt{3}}\frac{1}{1+x^2}{\sin }^{-1}\left(\frac{2x}{1+x^2}\right)dx$
Now ${\sin }^{-1}\left(\frac{2x}{1+x^2}\right)=\{\begin{array}{cc}2{\tan }^{-1}x,& \text{if}-1\le x\le 1\\ \pi -2{\tan }^{-1}x,& \text{if}x>1\end{array}$
$I={\int }_0^1\frac{1}{1+x^2}{\sin }^{-1}\left(\frac{2x}{1+x^2}\right)dx+{\int }_1^{\sqrt{3}}\frac{1}{1+x^2}{\sin }^{-1}\left(\frac{2x}{1+x^2}\right)dx$
$={\int }_0^1\frac{2{\tan }^{-1}x}{1+x^2}dx+{\int }_1^{\sqrt{3}}\frac{\pi -2{\tan }^{-1}x}{1+x^2}dx$
$=2{\int }_0^1\frac{{\tan }^{-1}x}{1+x^2}dx+\pi {\int }_1^{\sqrt{3}}\frac{1}{1+x^2}dx-2{\int }_1^{\sqrt{3}}\frac{{\tan }^{-1}x}{1+x^2}dx$
$=2{\int }_0^{\frac{\pi }{4}}tdt+\pi {\left({\tan }^{-1}x\right)}_1^{\sqrt{3}}-2{\int }_{\frac{\pi }{4}}^{\frac{\pi }{3}}tdt$
$=2{\left(\frac{t^2}{2}\right)}_0^{\frac{\pi }{4}}+\pi \{{\tan }^{-1}\sqrt{3}-{\tan }^{-1}1\}-2{\left(\frac{t^2}{2}\right)}_{\frac{\pi }{4}}^{\frac{\pi }{3}}$
$=\frac{{\pi }^2}{16}+\pi \{\frac{\pi }{3}-\frac{\pi }{4}\}-\{\frac{{\pi }^2}{9}-\frac{{\pi }^2}{16}\}=\frac{7}{72}{\pi }^2$
Hence, option 'C' is correct.