# The value of the integral

Question:

The value of the integral $\int_{-1}^{1} \log _{e}(\sqrt{1-x}+\sqrt{1+x}) d x$ is equal to :

1. $\frac{1}{2} \log _{\mathrm{e}} 2+\frac{\pi}{4}-\frac{3}{2}$

2. $2 \log _{e} 2+\frac{\pi}{4}-1$

3. $\log _{e} 2+\frac{\pi}{2}-1$

4. $2 \log _{e} 2+\frac{\pi}{2}-\frac{1}{2}$

Correct Option: , 3

Solution:

Let $I=2 \int_{0}^{1} \underbrace{\ln (\sqrt{1-x}+\sqrt{1+x})}_{\text {(I) }} 1 \mathrm{dx}$

(I.B.P.)

$\therefore \mathrm{I}=2\left[(\mathrm{x} \cdot \ln (\sqrt{1-\mathrm{x}}+\sqrt{1-\mathrm{x}}))_{0}^{1}\right.$

$\left.-\int_{0}^{1} x \cdot\left(\frac{1}{\sqrt{1-x}+\sqrt{1+x}}\right) \cdot\left(\frac{1}{2 \sqrt{1+x}}-\frac{1}{2 \sqrt{1-x}}\right) d x\right)$

$=2(\ln \sqrt{2}-0)-\frac{2}{2} \int_{0}^{1} \frac{x \sqrt{1-x}-\sqrt{1+x} d x}{(\sqrt{1-x}+\sqrt{1+x}) \sqrt{1-x^{2}}}$

$=\left(\log _{e} 2\right)-\int_{0}^{1} \frac{x \cdot\left(2-2 \sqrt{1-x^{2}}\right)}{-2 x \sqrt{1-x^{2}}} d x$

(After rationalisation)

$=\left(\log _{e} 2\right)+\int_{0}^{1}\left(\frac{1-\sqrt{1-x^{2}}}{\sqrt{1-x^{2}}}\right) d x$

$=\left(\log _{\mathrm{e}} 2\right)+\left(\sin ^{-1} \mathrm{x}\right)_{0}^{1}-1$

$=\log _{\mathrm{e}} 2+\left(\frac{\pi}{2}-0\right)-1$

$\therefore \quad I=\left(\log _{e} 2\right)+\frac{\pi}{2}-1$