Integrating the Area Under a Shaded Region Using Substitution

[kgcollegebound]

the problem asks for the area under the shaded region of the line y = 1/(1-x^2) on the interval [-1,1].

so far I've set up the integral showing
\int [tex]dx/(1-x^2)[\tex]
on the interval [-1,1]


i'm pretty sure you have to use substitution to solve it, but i can't seem to figure it out
please show/describe each step to solve it.

 

[sennyk]

I'll give you a hint. You will start with trig substitution.

 

[bob1182006]

$$\int \frac{dx}{1-x^2}=\int \frac{dx}{(1+x)(1-x)}$$

and do partial fractions, or you can do the trig substitution.

 

[Degoncire]

Well, to not get anyone confused, I'm taking out my calculations, cus its like 5 o clock in teh morning and i don't quite remember all the steps i was doing, so i might re attack this later , when I'm awake.

 

[Mute]

-1∫1 [ 1 / (1 - x^2) * dx ]

u = (1 - x^2)
dx = -2x

-1∫1 [ 1/u * dx ] = -1|1 [ ln(u) * dx ]

-1|1 [ -2x * ln(1 - x^2) ]

x=-1 [ 2ln(0) ]
x=1 [ -2ln(0) ]

ln(0) = infinity
[ 2ln(0) - -2ln(0) ] = +undefined or +infinity

(my knowledge does not go as far as to know what the end result is, i just know it's a positive end result.)

That is incorrect. You made the substitution $u = 1 - x^2$ in the integrand, but failed to make the change $dx \rightarrow -\frac{du}{2x}$ (and then substituting in the expression for x in terms of u, which is multivalued on [-1,1], so we'd want to reduce the integration range to [0,1] by noting the integrand is even).

Hence, the correct integral to evaluate, using that subsitution, is

$$2\int_1^0 du~\frac{1}{u\sqrt{1-u}}$$

which really isn't much better.

 

[Degoncire]

Darnit, your right, i was typing that in there after figuring it out, and i knew i did something wrong, but i couldn't figure it out. Thanks a lot.