Question

If $I={\int }_0^a\sqrt{\frac{a-x}{a+x}}dx,a>0,$ then $I$ equals

• A. $\frac{1}{2}(a-\frac{\pi }{2})$
• B. $\frac{a}{2}(\pi -1)$
• C. $\frac{1}{\sqrt{2}}a(\pi -1)$
• D. $a(\frac{\pi }{2}-1)$

Answer 1

The correct option is D $a(\frac{\pi }{2}-1)$

$I={\int }_0^a\sqrt{\frac{a-x}{a+x}}dx$
$I={\int }_0^a\sqrt{\frac{a-x}{a+x}\times \frac{a-x}{a-x}}dx$
$I={\int }_0^a\frac{a-x}{\sqrt{a^2-x^2}}dx$
$I={\int }_0^a\frac{a}{\sqrt{a^2-x^2}}dx-{\int }_0^a\frac{x}{\sqrt{a^2-x^2}}dx$
$I=a{[{\sin }^{-1}\frac{x}{a}]}_0^a-{\int }_0^a\frac{x}{\sqrt{a^2-x^2}}dx$
Substitute $a^2-x^2=t^2$
$\Rightarrow -xdx=tdt$
When $x=0\Rightarrow t=a$
When $x=a\Rightarrow t=0$
$I=a{\left[{\sin }^{-1}\frac{x}{a}\right]}_0^a-{\int }_0^a\frac{t}{\sqrt{t^2}}dt$
$I=a\frac{\pi }{2}-a$
$\Rightarrow I=a(\frac{\pi }{2}-1)$