Question

Evaluate ${\int }_{-1}^1\log \left[\sqrt{(1-x)}+\sqrt{(1+x)}\right]dx$

Answer 1

Step 1. Find the value of ${\mathbf{\int }}_{\mathbf{-}\mathbf{1}}^{\mathbf{1}}\mathit{l}\mathit{o}\mathit{g}\mathbf{}\left[\sqrt{(1-x)}+\sqrt{(1+x)}\right]\mathit{d}\mathit{x}$:

Let $I={\int }_{-1}^1\log \left[\sqrt{(1-x)}+\sqrt{(1+x)}\right]dx$

Let $x=\cos 2\theta$

differentiate it with respect to $\theta$

$dx=-2\sin 2\theta d\theta$

When $x=-1,\cos 2\theta =\cos \mathrm{\pi }$ $\left[\mathbf{\because }\mathbf{cos}\mathbf{}\mathbf{\pi }\mathbf{}\mathbf{=}\mathbf{}\mathbf{-}\mathbf{1}\right]$

$\Rightarrow$ $2\theta =\mathrm{\pi }$

$\Rightarrow$ $\theta =\frac{\mathrm{\pi }}{2}$

When $x=1,\cos 2\theta =\cos 0$ $\left[\mathbf{\because }\mathbf{cos}\mathbf{}\mathbf{0}\mathbf{}\mathbf{=}\mathbf{}\mathbf{1}\right]$

$\Rightarrow$ $2\theta =0$

$\Rightarrow$ $\theta =0$

Now, $I={\int }_{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}^0\ln \left[\sqrt{(1-\cos 2\theta )}+\sqrt{(1+\cos 2\theta )}\right]\left(-2\sin 2\theta d\theta \right)$

$\Rightarrow$ $I={\int }_{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}^0\ln \left(\sqrt{2{\sin }^2\theta }+\sqrt{2{\cos }^2\theta }\right)\left(-2\sin 2\theta \right)d\theta$

$\Rightarrow$ $I={\int }_{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}^0\ln \left(\sqrt{2}\sin \theta +\sqrt{2}\cos \theta \right)\left(-2\sin 2\theta \right)d\theta$

$\Rightarrow$ $I={\int }_{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}^0\ln \sqrt{2}\left(\sin \theta +\cos \theta \right)\left(-2\sin 2\theta \right)d\theta$

$\Rightarrow$ $I={\int }_{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}^0\left[\ln \sqrt{2}+\ln \left(\sin \theta +\cos \theta \right)\right]\left(-2\sin 2\theta \right)d\theta$ ; $\left[\mathbf{\because }\mathit{l}\mathit{o}\mathit{g}\left(ab\right)\mathbf{}\mathbf{=}\mathbf{}\mathit{l}\mathit{o}\mathit{g}\mathbf{}\mathit{a}\mathbf{}\mathbf{+}\mathbf{}\mathit{l}\mathit{o}\mathit{g}\mathbf{}\mathit{b}\right]$

$\Rightarrow$ $I={\int }_{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}^0\ln \sqrt{2}\left(-2\sin 2\theta \right)d\theta +{\int }_{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}^0\ln \left(\sin \theta +\cos \theta \right)\left(-2\sin 2\theta \right)d\theta$

$\Rightarrow$ $I=\ln \sqrt{2}{\int }_{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}^0\left(-2\sin 2\theta \right)d\theta +{\int }_{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}^0\ln \left(\sin \theta +\cos \theta \right)\left(-2\sin 2\theta \right)d\theta$

$\Rightarrow$ $I=\ln \sqrt{2}{\left[-2\left(\frac{-\cos 2\theta }{2}\right)\right]}_{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}^0-2{\int }_{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}^0\ln \left(\sin \theta +\cos \theta \right)\sin 2\theta d\theta$

$\Rightarrow$ $I=\ln \sqrt{2}{\left[\cos 2\theta \right]}_{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}^0-2{\int }_{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}^0\ln \left(\sin \theta +\cos \theta \right)\sin 2\theta d\theta$

Step 2. Apply the formula of integration by parts, we get

$\left[\mathbf{\because }\mathbf{\int }\left(u·v\right)\mathbf{d}\mathit{x}\mathbf{}\mathbf{=}\mathbf{}\mathit{u}\mathbf{\int }\mathit{v}\mathbf{d}\mathit{x}\mathbf{}\mathbf{-}\mathbf{\int }\left\{\frac{d\left(u\right)}{dx}.\int vdx\right\}\mathbf{d}\mathit{x}\right]$

$\Rightarrow$$I=\ln \sqrt{2}\left[\cos 0-\cos \mathrm{\pi }\right]-2\left[\ln \left(\sin \theta +\cos \theta \right)\int \sin 2\theta -\int \frac{d\ln \left(\sin \theta +\cos \theta \right)}{dx}.\int \sin 2\theta d\theta \right]$

$\Rightarrow$$I=\ln \sqrt{2}\left[1-\left(-1\right)\right]-2\left[\ln \left(\sin \theta +\cos \theta \right)\int \sin 2\theta -\int \frac{d\ln \left(\sin \theta +\cos \theta \right)}{dx}.\int \sin 2\theta d\theta \right]$

$\Rightarrow$$I=2\ln \sqrt{2}+I_1$

$I_1=-2\left[\ln \left(\sin \theta +\cos \theta \right)\int \sin 2\theta -\int \frac{d\ln \left(\sin \theta +\cos \theta \right)}{dx}.\int \sin 2\theta d\theta \right]$

$\Rightarrow$$I_1=-2{\left[\ln \left(\sin \theta +\cos \theta \right)\left(\frac{-\cos 2\theta }{2}\right)\right]}_{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}^0+2\left[\left(\frac{1}{\cos \theta +\sin \theta }\right)\times \left(-\sin \theta +\cos \theta \right)\times \left(\frac{-\cos 2\theta }{2}\right)d\theta \right]$

$\Rightarrow$$I_1=-2\left[\ln \left(\sin 0+\cos 0\right)\left(\frac{-\cos 0}{2}\right)-\ln \left(\sin \frac{\mathrm{\pi }}{2}+\cos \frac{\mathrm{\pi }}{2}\right)\left(\frac{-\cos {\displaystyle \raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{2}\right)\right]+2\left[\left(\frac{1}{\cos \theta +\sin \theta }\right)\times \left(-\sin \theta +\cos \theta \right)\times \left(\frac{-\cos 2\theta }{2}\right)d\theta \right]$

$\Rightarrow$$I_1=-2\left[\ln \left(1\right)\left(\frac{-1}{2}\right)-\ln \left(1+0\right)\left(\frac{1}{2}\right)\right]+2\left[\left(\frac{1}{\cos \theta +\sin \theta }\right)\times \left(-\sin \theta +\cos \theta \right)\times \left(\frac{-\cos 2\theta }{2}\right)d\theta \right]$

$\Rightarrow$$I_1=-2\left[0-0\right]+2\left[\left(\frac{1}{\cos \theta +\sin \theta }\right)\times \left(-\sin \theta +\cos \theta \right)\times \left(\frac{-\cos 2\theta }{2}\right)d\theta \right]$

$\Rightarrow$$I_1=0+2\left[\left(\frac{1}{\cos \theta +\sin \theta }\right)\times \left(-\sin \theta +\cos \theta \right)\times \left(\frac{-\cos 2\theta }{2}\right)d\theta \right]$

$\Rightarrow$$I_1=2\left[{\int }_{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}^0\left(\frac{\cos \theta -\sin \theta }{\cos \theta +\sin \theta }\right)\left(\frac{-1}{2}\cos 2\theta \right)d\theta \right]$

$\Rightarrow$$I_1=-2\times -\frac{1}{2}\left[{\int }_0^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\left(\frac{\cos \theta -\sin \theta }{\cos \theta +\sin \theta }\right)\left(\cos 2\theta \right)d\theta \right]$

$\Rightarrow$$I_1={\int }_0^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\left(\frac{\cos \theta -\sin \theta }{\cos \theta +\sin \theta }\right)\left({\cos }^2\theta -{\sin }^2\theta \right)d\theta$

$\Rightarrow$$I_1={\int }_0^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{\left(\cos \theta -\sin \theta \right)}^{2}d\theta$ $\left[\mathbf{\because }{\mathit{a}}^{\mathbf{2}}\mathbf{}\mathbf{-}\mathbf{}{\mathit{b}}^{\mathbf{2}}\mathbf{}\mathbf{=}\mathbf{}\left(a+b\right)\left(a-b\right)\right]$

$\Rightarrow$$I_1={\int }_0^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\left({\cos }^2\theta +{\sin }^2\theta -2\sin \theta \cos \theta \right)d\theta$ $\left[\mathbf{\because }{\left(a-b\right)}^{\mathbf{2}}\mathbf{}\mathbf{=}\mathbf{}{\mathit{a}}^{\mathbf{2}}\mathbf{}\mathbf{+}\mathbf{}{\mathit{b}}^{\mathbf{2}}\mathbf{}\mathbf{-}\mathbf{}\mathbf{2}\mathit{a}\mathit{b}\right]$

$\Rightarrow$$I_1={\int }_0^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\left(1-\sin 2\theta \right)d\theta$ $\left[\mathbf{\because }\mathbf{sin}\mathbf{}\mathbf{2}\mathit{\theta }\mathbf{}\mathbf{=}\mathbf{}\mathbf{2}\mathbf{}\mathbf{sin}\mathbf{}\mathit{\theta }\mathbf{cos}\mathbf{}\mathit{\theta }\right]$

$\Rightarrow$$I_1={\int }_0^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}d\theta -{\int }_0^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\sin 2\theta d\theta$

$\Rightarrow$$I_1={\left[\theta \right]}_0^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-{\left[\frac{-\cos 2\theta }{2}\right]}_0^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

$\Rightarrow$$I_1=\left[\frac{\mathrm{\pi }}{2}-0\right]+\frac{1}{2}\left[\cos \mathrm{\pi }-\cos 0\right]$

$\Rightarrow$$I_1=\frac{\mathrm{\pi }}{2}+\frac{1}{2}\left[-1-1\right]$

$\Rightarrow$$I_1=\frac{\mathrm{\pi }}{2}-1$

Step 3. Put the value of $I_1$ in $I=2\ln \sqrt{2}+I_1$, we get

$I=2\ln \sqrt{2}+I_1$

$\Rightarrow$$I=2\ln \sqrt{2}+\frac{\mathrm{\pi }}{2}-1$

$\Rightarrow$$I=2\ln \sqrt{2}-1+\frac{\mathrm{\pi }}{2}$

Hence, ${\mathbf{\int }}_{\mathbf{-}\mathbf{1}}^{\mathbf{1}}\mathit{l}\mathit{o}\mathit{g}\mathbf{}\mathbf{[}\sqrt{\mathbf{(}\mathbf{1}\mathbf{}\mathbf{-}\mathbf{}\mathbf{x}\mathbf{)}\mathbf{}}\mathbf{+}\mathbf{}\sqrt{\mathbf{(}\mathbf{1}\mathbf{}\mathbf{+}\mathbf{}\mathbf{x}\mathbf{)}}\mathbf{]}\mathit{d}\mathit{x}\mathbf{}\mathbf{=}\mathbf{}\mathbf{2}\mathit{l}\mathit{n}\sqrt{\mathbf{2}}\mathbf{}\mathbf{-}\mathbf{}\mathbf{1}\mathbf{}\mathbf{+}\mathbf{}\frac{\mathbf{\pi }}{\mathbf{2}}$.