Question

If ${\int }_0^1{x}^5\sqrt{\frac{1+x^2}{1-x^2}}dx=m\pi +n$, then the ordered pair $(m,n)$ is equal to:

• A. $(\frac{1}{3},\frac{1}{8})$
• B. $(\frac{1}{8},\frac{2}{3})$
• C. $(\frac{1}{4},\frac{1}{3})$
• D. $(\frac{1}{8},\frac{1}{3})$

Answer 1

The correct option is D $(\frac{1}{8},\frac{1}{3})$

$\begin{array}{c}Given,\\ I={\int }_0^1{x}^5\sqrt{\frac{1+x^2}{1-x^2}}\\ Let{x}^2=tthen,2xdu=dt\Rightarrow xdx=\frac{dt}{2}\\ Hence,\\ I={\int }_0^1{t}^2\sqrt{\frac{1+t}{1-t}}\frac{dt}{2}\\ =\frac{1}{2}{\int }_0^1\frac{t^2\left(1+t\right)}{\sqrt{1-t^2}}dt\\ =\frac{1}{2}{\int }_0^1\frac{t^2}{\sqrt{1-t^2}}dt+\frac{1}{2}{\int }_2^1\frac{t^3}{\sqrt{1-t^2}}dt\\ Lett=\sin \theta thendt=\cos \theta d\theta \\ =\frac{1}{2}{\int }_0^{\frac{\pi }{2}}{\sin }^2\theta d\theta +\frac{1}{2}{\int }_0^{\frac{\pi }{2}}\left(1-{\cos }^{2}x\right)\sin \theta d\theta \\ =\frac{1}{4}{\int }_0^{\frac{\pi }{2}}\left(1-\cos 2\theta \right)d\theta +\frac{1}{2}{\int }_0^{\frac{\pi }{2}}-\cos \theta d\theta +\frac{1}{2}{\int }_0^{\frac{\pi }{2}}\frac{{\cos }^3\theta }{3}d\theta \\ =\frac{1}{2}{\int }_0^{\frac{\pi }{2}}\theta d\theta -{\int }_0^{\frac{\pi }{2}}\sin 2\theta d\theta +\frac{1}{2}-\frac{1}{2}+\frac{1}{3}\\ =\frac{\pi }{8}+\frac{1}{2}-\frac{1}{6}\\ =\frac{\pi }{8}+\frac{3-1}{6}\\ =\frac{\pi }{8}+\frac{1}{3}\\ Now,m\pi +n=\left(\frac{1}{8},\frac{1}{3}\right)\\ Hence,theoptionDisthecorrectanswer.\end{array}$