Definite Logarithm

We start by using the logarithmic identity $\log_b (x^d) = d \log_b (x)$ . Using this, we can write the integrand as $\frac{1}{2} \ln \left( \sqrt{1 - x} + \sqrt{1 + x} \right)^2$ . Thus, the integral becomes

$\begin{aligned} I &= \int_0^1 \ln \left( \sqrt{1 - x} + \sqrt{1 + x} \right) \, dx \\ &= \int_0^1 \frac{1}{2} \ln \left( \sqrt{1 - x} + \sqrt{1 + x} \right)^2 \, dx \\ &= \int_0^1 \frac{1}{2} \ln \left( (1 - x) + 2\sqrt{(1 - x)(1 + x)} + (1 - x) \right) \, dx \\ &= \int_0^1 \frac{1}{2} \ln \left( 2 + 2\sqrt{1 - x^2} \right) \, dx \end{aligned}$

From here, we make the substitution $x = \sin \theta$ :

$\begin{aligned} I &= \int_0^1 \frac{1}{2} \ln \left( 2 + 2\sqrt{1 - x^2} \right) \, dx \\ &= \int_0^{\pi / 2} \frac{1}{2} \ln \left( 2 + 2\sqrt{1 - \sin^2 \theta} \right) (cos \theta \, d \theta) \\ &= \int_0^{\pi / 2} \frac{1}{2} \ln \left( 2 + 2 \cos \theta \right) cos \theta \, d \theta \end{aligned}$

Now, we use integration by parts, such that

$\begin{aligned} u &= \ln (2 + 2 \cos \theta) & dv &= \cos \theta \, d \theta \\ du &= \frac{-2 \sin \theta}{2 + 2 \cos \theta} \, d \theta & v &= \sin \theta \end{aligned}$

Then, the value of the integral becomes

$\begin{aligned} I &= \int_0^{\pi / 2} \frac{1}{2} \ln \left( 2 + 2 \cos \theta \right) cos \theta \, d \theta \\ &= \frac{1}{2} \left( \left. (\ln (2 + 2 \cos \theta) \sin \theta) \right|_0^{\pi / 2} - \int_0^{\pi / 2} \frac{-2 \sin^2 \theta}{2 + 2 \cos \theta} \, d \theta \right) \\ &= \frac{1}{2} \left( \ln (2) + \int_0^{\pi / 2} \frac{\sin^2 \theta}{1 + \cos \theta} \, d \theta \right) \end{aligned}$

The resulting integral on the right can be simplified using the trigonometric identity $\sin^2 \theta + \cos^2 \theta = 1$ . Substituting $1 - \cos^2 \theta$ for $\sin^2 \theta$ , we see that

$\begin{aligned} \int_0^{\pi / 2} \frac{\sin^2 \theta}{1 + \cos \theta} \, d \theta &= \int_0^{\pi / 2} \frac{1 - \cos^2 \theta}{1 + \cos \theta} \, d \theta \\ &= \int_0^{\pi / 2} \frac{(1 - \cos \theta)(1 + \cos \theta)}{1 + \cos \theta} \, d \theta \\ &= \int_0^{\pi / 2} (1 - \cos \theta) \, d \theta \\ &= \int_0^{\pi / 2} 1 \, d \theta - \int_0^{\pi / 2} \cos \theta \, d \theta \\ &= \frac{\pi}{2} - 1 \end{aligned}$

Thus, the value of the integral is

$\begin{aligned} I &= \frac{1}{2} \left( \ln (2) + \int_0^{\pi / 2} \frac{\sin^2 \theta}{1 + \cos \theta} \, d \theta \right) \\ &= \frac{1}{2} \left( \ln (2) - 1 + \frac{\pi}{2} \right) \\ &= \frac{1}{2} \ln (2) - \frac{1}{2} + \frac{\pi}{4} \end{aligned}$

And therefore,

$A + B + C + D = 2 + 2 + 2 + 4 = \boxed{10}$

Solution IMGUR

$Let\quad x=\sin { 2\theta } \quad for\quad \theta \in \left( 0,\cfrac { \pi }{ 4 } \right) \\ \\ \Rightarrow \sqrt { 1-x } =\left| \sin { \theta } -\cos { \theta } \right| =\cos { \theta } -\sin { \theta } \quad for\quad \theta \in \left( 0,\cfrac { \pi }{ 4 } \right) \\ \\ similarily,\sqrt { 1+x } =\cos { \theta } +\sin { \theta } \\ \\ \Rightarrow \sqrt { 1-x } +\sqrt { 1+x } =2\cos { \theta } \\ \\ dx=2\cos { 2\theta \quad d\theta } \\ \\ \therefore I=\int _{ 0 }^{ 1 }{ \ln { \left( \sqrt { 1-x } +\sqrt { 1+x } \right) } } dx=\int _{ 0 }^{ \cfrac { \pi }{ 4 } }{ \ln { \left( 2\cos { \theta } \right) } } .2\cos { 2\theta \quad d\theta } \\ \\ by\quad parts,\int { \ln { \left( 2\cos { \theta } \right) } } .2\cos { 2\theta \quad d\theta } =\sin { 2\theta .\ln { \left( 2\cos { \theta } \right) } } -\int { \sin { 2\theta \cfrac { -2\sin { \theta } }{ 2\cos { \theta } } } } d\theta \\ \\ =\sin { 2\theta .\ln { \left( 2\cos { \theta } \right) } } +\int { 2\sin ^{ 2 }{ \theta } } d\theta =\sin { 2\theta .\ln { \left( 2\cos { \theta } \right) } } -\cfrac { 1 }{ 2 } \sin { 2\theta } +\theta +C\\ \\ \therefore \int _{ 0 }^{ \cfrac { \pi }{ 4 } }{ \ln { \left( 2\cos { \theta } \right) } } .2\cos { 2\theta \quad d\theta } ={ \left[ \sin { 2\theta .\ln { \left( 2\cos { \theta } \right) } } -\cfrac { 1 }{ 2 } \sin { 2\theta } +\theta \right] }_{ 0 }^{ \cfrac { \pi }{ 4 } }\\ \\ =\cfrac { 1 }{ 2 } \ln { 2 } -\cfrac { 1 }{ 2 } +\cfrac { \pi }{ 4 } \\ \\ \therefore A+B+C+D=2+2+2+4=10$

Just try using normal integration by parts without substitution to get a simple and elegant solution!!