how to integrate [log({sqrt(1-x)}+{sqrt(1+x)}] where lower limit is 0 and upper limit is 1

Tuesday, January 22, 2019

how to integrate [log({sqrt(1-x)}+{sqrt(1+x)}] where lower limit is 0 and upper limit is 1

$\displaystyle{{\left(\sqrt{{{1}-{x}}}+\sqrt{{{1}+{x}}}\right)}}^{{2}}$ />
$\displaystyle={1}-{x}+{1}+{x}+{2}\sqrt{{{\left({1}-{x}\right)}{\left({1}+{x}\right)}}}$
/> $\displaystyle={2}{\left({1}+\sqrt{{{1}-{{x}}^{{2}}}}\right)}$

So:
/> $\displaystyle\text{log}$
(sqrt(1-x)+sqrt(1+x))

$\displaystyle=$

"log"((2(1+sqrt(1-x^2)))^(1/2))

$\displaystyle=\frac{{{1}}}{{{2}}}{\left(\text{log}{2}+\right.}$

"log" (1+sqrt(1-x^2) ))

So:
/> $\displaystyle\int$
("log" (sqrt(1-x)+sqrt(1+x)) ) dx

$\displaystyle=\frac{{{1}}}{{{2}}}$
int ("log"
2) dx + (1)/(2) int ("log" (1+sqrt(1-x^2))) dx
/>

The first integral is just

constant:

$\displaystyle\frac{{{1}}}{{{2}}}{\int_{{0}}^{{1}}}{\left(\text{log}{2}\right)}{\left.{d}{x}\right.}=\frac{{{1}}}{{{2}}}$
("log"
2) (1-0) = ("log" 2)/(2)

/>For the second
integral, we start with:

$\displaystyle{\int_{{0}}^{{1}}}$
("log" (1+sqrt(1-x^2)))dx

We start with a trig
substitution:

$\displaystyle{x}=\text{sin}$
theta
/> $\displaystyle{\left.{d}{x}\right.}=\text{cos}\theta{d}\theta$

$\displaystyle{1}-{{x}}^{{2}}={1}-$
"sin"^2
theta = "cos"^2 theta

$\displaystyle\sqrt{{{1}-{{x}}^{{2}}}}=$
"cos"
theta

If $\displaystyle\theta={0}$ then $\displaystyle{x}={0}$

If
$\displaystyle\theta$
= (pi)/(2) $\displaystyle{t}{h}{e}{n}$ x=1

/>And the integral
becomes:

$\displaystyle{\int_{{0}}^{{\frac{\pi}{{2}}}}}{\left(\text{log}\right.}$
(1+ "cos" theta)
"cos" theta) d theta

Now we use
integration by
parts:

$\displaystyle{u}=\text{log}{\left({1}+\text{cos}\right.}$

theta)

$\displaystyle{d}{v}=\text{cos}\theta{d}$
theta
/>So:

$\displaystyle{d}{u}=\frac{{-\text{sin}\theta}}{{{1}+}}$
"cos" theta) d
theta

$\displaystyle{v}=\text{sin}$
theta

/> $\displaystyle\int{\left(\text{log}{\left({1}+\text{cos}\theta\right)}\right.}$
"cos" theta) d theta
/> $\displaystyle=\text{log}{\left({1}+\text{cos}\theta\right)}$
"sin" theta - int (("sin" theta)(-"sin"
theta)/(1 +
"cos" theta)) d theta

$\displaystyle=\text{log}{\left({1}+\right.}$
"cos"
theta) "sin" theta + int (("sin"^2 theta)/(1 + "cos" theta))
d
theta

Now: />
$\displaystyle{\left({\text{sin}}^{{2}}\right.}$
theta)/(1+"cos" theta)

$\displaystyle=$
(("sin"^2
theta)(1-"cos" theta))/((1+"cos" theta)(1-"cos"

theta))

$\displaystyle={\left({\left({\text{sin}}^{{2}}\theta\right)}{\left({1}-\text{cos}\right.}\right.}$

theta))/(1-"cos"^2 theta)

$\displaystyle={\left({\left({\text{sin}}^{{2}}\right.}\right.}$
theta)(1-"cos"
theta))/("sin"^2
theta)

$\displaystyle={1}-\text{cos}\theta$ />

Thus our
integral becomes:
/> $\displaystyle=\text{log}{\left({1}+\text{cos}\theta\right)}$
"sin" theta + int (1-"cos" theta) d
theta

$\displaystyle=$
"log" (1+ "cos" theta) "sin" theta + theta -
"sin" theta
|_0^(pi/2)

/>
$\displaystyle{\int_{{0}}^{{1}}}{\left(\text{log}\right.}$
(sqrt(1-x)+sqrt(1+x)) ) dx />
$\displaystyle=\frac{{\text{log}{2}}}{{{2}}}+\frac{{{1}}}{{{2}}}{(}$
"log" (1+ "cos" theta) "sin"
theta + theta - "sin" theta
|_0^(pi/2) )

$\displaystyle={\left(\text{log}\right.}$
2)/(2) + (1)/(2)( "log" (1+ 0)
*1 + (pi)/(2) - 1 ) - (1)/(2)( "log" (1+ 1) *0
+ 0 - 0 )

$\displaystyle=$
("log" 2)/(2) + (1)/(2)( (pi)/(2) - 1
)

/>