`(sqrt(1-x)+sqrt(1+x))^2`
/>
`=1-x+1+x+2sqrt((1-x)(1+x))`
/>`=2(1+sqrt(1-x^2))`
So:
/>`"log"
(sqrt(1-x)+sqrt(1+x))`
`=
"log"((2(1+sqrt(1-x^2)))^(1/2))`
`= (1)/(2) ("log" 2 +
"log" (1+sqrt(1-x^2) ))`
So:
/>`int
("log" (sqrt(1-x)+sqrt(1+x)) ) dx`
`=(1)/(2)
int ("log"
2) dx + (1)/(2) int ("log" (1+sqrt(1-x^2))) dx
`
/>
The first integral is just
constant:
`(1)/(2) int_0^1 ("log" 2) dx = (1)/(2)
("log"
2) (1-0) = ("log" 2)/(2)`
/>For the second
integral, we start with:
`int_0^1
("log" (1+sqrt(1-x^2)))dx
`
We start with a trig
substitution:
`x = "sin"
theta`
/>`dx = "cos" theta d theta`
`1-x^2 = 1 -
"sin"^2
theta = "cos"^2 theta`
`sqrt(1-x^2) =
"cos"
theta`
If `theta = 0` then `x=0`
If
`theta
= (pi)/(2)` then `x=1`
/>And the integral
becomes:
`int_0^(pi/2) ("log"
(1+ "cos" theta)
"cos" theta) d theta`
Now we use
integration by
parts:
`u = "log" (1+ "cos"
theta)`
`dv = "cos" theta d
theta`
/>So:
`du = (-"sin" theta)/(1 +
"cos" theta) d
theta`
`v="sin"
theta`
/>`int ("log" (1+ "cos" theta)
"cos" theta) d theta`
/>`= "log" (1+ "cos" theta)
"sin" theta - int (("sin" theta)(-"sin"
theta)/(1 +
"cos" theta)) d theta`
`= "log" (1+
"cos"
theta) "sin" theta + int (("sin"^2 theta)/(1 + "cos" theta))
d
theta`
Now:
/>
`("sin"^2
theta)/(1+"cos" theta)`
`=
(("sin"^2
theta)(1-"cos" theta))/((1+"cos" theta)(1-"cos"
theta))`
`= (("sin"^2 theta)(1-"cos"
theta))/(1-"cos"^2 theta)`
`= (("sin"^2
theta)(1-"cos"
theta))/("sin"^2
theta)`
`=1-"cos" theta`
/>
Thus our
integral becomes:
/>`= "log" (1+ "cos" theta)
"sin" theta + int (1-"cos" theta) d
theta`
`=
"log" (1+ "cos" theta) "sin" theta + theta -
"sin" theta
|_0^(pi/2)`
/>
`int_0^1 ("log"
(sqrt(1-x)+sqrt(1+x)) ) dx`
/>
` = ("log" 2)/(2) + (1)/(2)(
"log" (1+ "cos" theta) "sin"
theta + theta - "sin" theta
|_0^(pi/2) )`
` = ("log"
2)/(2) + (1)/(2)( "log" (1+ 0)
*1 + (pi)/(2) - 1 ) - (1)/(2)( "log" (1+ 1) *0
+ 0 - 0 )`
` =
("log" 2)/(2) + (1)/(2)( (pi)/(2) - 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
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