^ _0 1-x 1+x dx= 1. 1- ^2-1 2. 2 -1 3. 2 +1 4. + 1 - Vidyadip

Question

$\int\limits^\pi_0\sqrt{\frac{1-\text{x}}{1+\text{x}}}\text{ dx}=$

$\sqrt{1-\pi^2}-1$
$\frac{\pi}{2}-1$
$\frac{\pi}{2}+1$
${\pi}+{1}$

✓

Answer

$\sqrt{1-\pi^2}-1$

Solution:
We have,
$\text{I}=\int\limits^\pi_0\sqrt{\frac{1-\text{x}}{1+\text{x}}}\text{dx}$
$=\int\limits^\pi_0\Big[\sqrt{\frac{1-\text{x}}{1+\text{x}}}\times\frac{\sqrt{1-\text{x}}}{\sqrt{1-\text{x}}}\Big]\text{dx}$
$=\int\limits^\pi_0\frac{1-\text{x}}{\sqrt{1-\text{x}^2}}\text{dx}$
$=\int\limits^\pi_0\frac{1}{\sqrt{1-\text{x}^2}}\text{dx}-\int\limits^\pi_0\frac{\text{x}}{\sqrt{1-\text{x}^2}}\text{dx}$
Putting $1 -\text{x}^2=\text{t}$
$\Rightarrow - 2\text{x}\text{ dx}=\text{dt}$
$\Rightarrow \text{x}\text{ dx}=\frac{-\text{dt}}{2}$
when $\text{x}\rightarrow0;\text{t}\rightarrow1$
and $\text{x}\rightarrow\pi;\text{ t}\rightarrow-\pi^2$
$\therefore\ \text{I}=\int\limits^\pi_0\frac{1}{\sqrt{1-\text{x}^2}}-\int\limits^{(1-\pi^2)}_1\frac{-\text{dt}}{2\sqrt{\text{t}}}$
$=\big[\sin^{-1}\text{x}\big]^{\pi}_0+\frac{2}{2}\big[\sqrt{\text{t}}\big]^{1-\pi^2}_1$
$=\big[0-0\big]+\Big[\sqrt{1-\pi^2}-\sqrt{1}\Big]$
$=\sqrt{1-\pi^2}-1$

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