Find the Integral Using Substitution: \int \frac{-2 \sqrt{1-x}}{2 + \sqrt{1-x}}

[thomas49th]

Homework Statement

By making the substituion t = \sqrt{1-x}

find \int \frac{1}{2 + \sqrt{1 - x}}

Homework Equations

The Attempt at a Solution

So t = (1-x)^\frac{1/2}
t' = - \frac{1}{2} (1 - x)^{-\frac{1}{2}}

dx = -2 \sqrt{1-x} dt

\int \frac{-2 \sqrt{1-x}}{2 + \sqrt{1-x}} dt

\int \frac{-2 \sqrt{1-x}}{2 + t} dt

But am I anywhere useful? Am I allowed to say

\int \frac{-2t}{2 + t} dt

because I've made the substation already? In that case it's a simple 2 ln|2+ \sqrt{1-x}|

But that is wrong as the answer is a nasty:

4ln|2+ \sqrt{1-x}| - 2 \sqrt{1-x} + c

Thanks
Thomas

 

[Gib Z]

But am I anywhere useful? Am I allowed to say

\int \frac{-2t}{2 + t} dt

because I've made the substation already? In that case it's a simple [tex 2 ln|2+ \sqrt{1-x}[/tex]

But that is wrong as the answer is a nasty:

4ln|2+ \sqrt{1-x}| - 2 \sqrt{1-x} + c

Thanks
Thomas

The integral you reached is correct, but You integrated incorrectly. Can you see where you went wrong?

 

[thomas49th]

The integral you reached is correct, but You integrated incorrectly. Can you see where you went wrong?

So

<br /> \int \frac{-2t}{2 + t} dt <br />

is right? I'm allowed to say t = \sqrt{1-x} when I change the dx to a t = \sqrt{1-x}dt.?

If so I can see why I can't use ln - because the deravative of the bottom is NOT on the top! Should I integrate by parts?

Thanks
Thomas

 

[Gib Z]

So

<br /> \int \frac{-2t}{2 + t} dt <br />

is right? I'm allowed to say t = \sqrt{1-x} when I change the dx to a t = \sqrt{1-x}dt.?
[\QUOTE]

I don't really understand what the last part of the sentence means, or the source of your confusion. Looking back at the first post, the ONLY difference between this integral and the step before it is the direct replacement of sqrt(1-x) with t, which is the substitution used! Where is the complication arising from?

If so I can see why I can't use ln - because the deravative of the bottom is NOT on the top! Should I integrate by parts?

You could use parts, but the two easier ways would be either 1) Subtract and add 4 into the numerator and split up the fraction, and simplify, or 2) Use a very easy substitution.

 

[Mark44]

You started with t = sqrt(1 - x) ==> t² = 1 - x ==> 2tdt = -dx.

With this substitution, you can change the original integral wholesale to this one:
\int \frac{-2t}{2 + t} dt

Do not use integration by parts here. Your integrand is an improper rational function (the degree of the numerator = the degree of the denominator). Divide the numerator by the denominator to get a simple polynomial and a proper rational function, then integrate.

Don't forget to undo your substitution.

Also - don't forget the dx, which you omitted at the start. Leaving off the differential term can come back around and bite you in more complicated problems.

 

[thomas49th]

You started with t = sqrt(1 - x) ==> t² = 1 - x ==> 2tdt = -dx.

With this substitution, you can change the original integral wholesale to this one:
\int \frac{-2t}{2 + t} dt

Do not use integration by parts here. Your integrand is an improper rational function (the degree of the numerator = the degree of the denominator). Divide the numerator by the denominator to get a simple polynomial and a proper rational function, then integrate.

Don't forget to undo your substitution.

Also - don't forget the dx, which you omitted at the start. Leaving off the differential term can come back around and bite you in more complicated problems.

Sorry I'm a little way off the answer. The substitution t^{2} = 1 - x

so you've implicitly differentiated that to 2tdt/dx = -1 right

\int \frac{-2t}{2 + t} dt

Divide the numerator by the denominator to get a simple polynomial and a proper rational function, then integrate.

I'm probably just being a complete spanner but the numerator is already divided by tge denominator. Are you trying to get something like

ax^{2} + bx + \frac{c}{2+t}

Sorry for the stupidity

Thomas

 

[Mark44]

Sorry I'm a little way off the answer. The substitution t^{2} = 1 - x

so you've implicitly differentiated that to 2tdt/dx = -1 right

Actually, I just took the differential of each side. If you multiply your equation by dx, you get my equation.

\int \frac{-2t}{2 + t} dt



I'm probably just being a complete spanner but the numerator is already divided by tge denominator. Are you trying to get something like

ax^{2} + bx + \frac{c}{2+t}

No, I'm trying to get an integrand that isn't an improper rational expression (an improper rational expression is one where the degree of the numerator is >= the degree of the denominator). For this integrand, the degree of the numerator is 1, and the degree of the denominator is 1. Divide the numerator by the denominator using polynomial long division. You should get -2 + <some number>/(2 + t).

Sorry for the stupidity

Thomas

 

[thomas49th]

Right. Gotcha

-2 + 4(t+2)

this integrates to
-2t + 4ln|t+2|

wap in the subst

4 ln |2 + \sqrt{1-x} | - 2\sqrt{1-x} +c

It's easy!
Thanks!