Question

Let $I={\int }_0^1\sqrt{\frac{1+\sqrt{x}}{1-\sqrt{x}}}dxandJ={\int }_0^1\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}dx$. Then which of the following statement(s) is/are correct?

• A. $I+J=4$
• B. $I-J=\frac{\pi }{2}$
• C. $I+J=2$
• D. $I-J=\pi$

Answer 1

Answer: (A), (D)

The correct options are
A $I+J=4$
D $I-J=\pi$

$I+J=\int \frac{(1+\sqrt{x})+(1-\sqrt{x})}{\sqrt{1-x}}dx$
$={\int }_0^1\frac{2}{\sqrt{1-x}}dx=2\times [-2\sqrt{1-x}{]}_0^1=4$
$I-J={\int }_0^1\frac{2\sqrt{x}}{\sqrt{1-x}}dx$ put $x=si{n}^2\theta dx=2sin\theta cos\theta d\theta$
$I-J={\int }_0^{\frac{\pi }{2}}\frac{2sin\theta .2sin\theta cos\theta d\theta }{cos\theta }$
$=4{\int }_0^{\frac{\pi }{2}}si{n}^2\theta d\theta =4{\int }_0^{\frac{\pi }{2}}\frac{1-cos2\theta }{2}d\theta$
$=2{[\theta -\frac{sin2\theta }{2}]}_0^{\frac{\pi }{2}}=\pi$