Question

If ${\int }_{-1/\sqrt{3}}^{1/\sqrt{3}}\frac{x^4}{1-x^4}{\cos }^{-1}\frac{{2x}^{}}{1{+x}^2}dx=k{\int }_0^{1/\sqrt{3}}\frac{x^4}{1-x^4}$ then $k$ equals

• A. $\pi$
• B. $2\pi$
• C. 2
• D. 1

Answer 1

Answer: (A)

$\pi$

${\int }_{-1/\sqrt{3}}^{1/\sqrt{3}}\frac{x^4}{1-x^4}{\cos }^{-1}\frac{2x}{1+x^2}dx$

This can be written as

$\Rightarrow {\int }_{-1/\sqrt{3}}^{1/\sqrt{3}}\frac{x^4}{1-x^4}(\frac{\pi }{2}-{\sin }^{-1}\frac{2x}{1+x^2})dx$

$\Rightarrow {\int }_{-1/\sqrt{3}}^{1/\sqrt{3}}\frac{x^4}{1-x^4}\frac{\pi }{2}dx-{\int }_{-1/\sqrt{3}}^{1/\sqrt{3}}{\sin }^{-1}\frac{2x}{1+x^2}dx$

Note that second integral is odd function. Hence it is zero

$\Rightarrow \frac{\pi }{2}{\int }_{-1/\sqrt{3}}^{1/\sqrt{3}}\frac{x^4}{1-x^4}dx$

$\Rightarrow \frac{\pi }{2}\times 2{\int }_0^{1/\sqrt{3}}\frac{x^4}{1-x^4}dx$

$\Rightarrow \pi {\int }_0^{1/\sqrt{3}}\frac{x^4}{1-x^4}dx$

$\Rightarrow k=\pi$