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The substitution u=tan(x/2)

  • Thread starter noblerare
  • Start date
47
0
I'm looking to solve this integral with the tan(x/2) substitution but so far, I don't know what to do.

1. Homework Statement

[tex]\int[/tex][tex]\sqrt{1-sinx}dx[/tex]

2. Homework Equations

u=tan(x/2)

3. The Attempt at a Solution

Well, using the tan(x/2) substitution, I get that:

sinx=[tex]\frac{2u}{1+u^2}[/tex]
dx=[tex]\frac{2du}{1+u^2}[/tex]

So I get:
[tex]\int\frac{2(u-1)du}{(1+u^2)^(3/2)}[/tex]

Now I don't know what to do.

Another method I tried is multiplying by conjugate:

[tex]\int[/tex][tex]\sqrt{1-sinx}[/tex][tex]\frac{\sqrt{1+sinx}}{\sqrt{1+sinx}}[/tex]dx

[tex]\int[/tex][tex]\frac{\sqrt{1-sin^2(x)}}{\sqrt{1+sinx}}[/tex]dx

[tex]\int[/tex][tex]\frac{cosxdx}{\sqrt{1+sinx}}[/tex]dx

u=sinx
du=cosxdx

[tex]\int[/tex][tex]\frac{du}{\sqrt{1+u}}[/tex]

[tex]\int[/tex](u+1)^(-[tex]\frac{1}{2}[/tex])du

2(u+1)^([tex]\frac{1}{2}[/tex]

2[tex]\sqrt{sinx+1}[/tex]+ C

Obviously, this is the wrong answer but I don't see where I went wrong.

Which way should I approach this problem?
 

Answers and Replies

tiny-tim
Science Advisor
Homework Helper
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sinx=[tex]\frac{2u}{1+u^2}[/tex]
dx=[tex]\frac{2du}{1+u^2}[/tex]

So I get:
[tex]\int\frac{2(u-1)du}{(1+u^2)^(3/2)}[/tex]
Hi noblerare! :smile:

Actually, you've gone too far.

tan(x/2) can be useful just for simplifying a formula … you don't necessarily have to change the vatiable of integration also.

In this case, just stop at √(1 - sinx) = √(1 - u)^2/√(1 + u^2).

Now translate it back to x … it's cos(x/2) - sin(x/2).

(to check, note that 1 - sinx = cos^2(x/2) - 2cos(x/2)sin(x/2) + sin^2(x/2) !).

So ∫√(1 - sinx)dx = √(cos(x/2) - sin(x/2))dx. :smile:
 
1,631
4
}}[/tex]dx

[tex]\int[/tex][tex]\frac{\sqrt{1-sin^2(x)}}{\sqrt{1+sinx}}[/tex]dx

[tex]\int[/tex][tex]\frac{cosxdx}{\sqrt{1+sinx}}[/tex]dx



2[tex]\sqrt{sinx+1}[/tex]+ C

Obviously, this is the wrong answer but I don't see where I went wrong.

Which way should I approach this problem?
Why do u think that this is the wrong answer, it looks fine to me?

[tex](2\sqrt{sinx+1}+C)'=2\frac{1}{2}\frac{cosx}{\sqrt{1+sinx}}=\frac{cosx}{\sqrt{1+sinx}}*\frac{\sqrt{1-sinx}}{\sqrt{1-sinx}}=\frac{cosx\sqrt{1-sinx}}{\sqrt{1-sin^{2}x}}=\frac{cosx\sqrt{1-sinx}}{cosx}=\sqrt{1-sinx}[/tex]

If the answer in the book is different from this one, they should be equivalent, because nothing is wrong with this.
 
Last edited:
47
0
I'm sorry, tiny-tim, I still don't quite understand what you're trying to say

I understand this
In this case, just stop at √(1 - sinx) = √(1 - u)^2/√(1 + u^2).
So far, I left my integral as so...

but
Now translate it back to x … it's cos(x/2) - sin(x/2).
how do I change it back to x? I also don't see how cos^2(x/2) -2cos(x/2)sin(x/2) + sin^2(x/2) comes out to 1-sinx
 
1,631
4
Noblerare,

Your answer is correct, what else are you looking for??
 
tiny-tim
Science Advisor
Homework Helper
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249
… standard trig formulas …

how do I change it back to x?
Because √(1 - u)^2/√(1 + u^2) = (1 - u)/√(1 + u^2)
= (1 - tan(x/2))cos(x/2)
= cos(x/2) - sin(x/2). :smile:
I also don't see how cos^2(x/2) -2cos(x/2)sin(x/2) + sin^2(x/2) comes out to 1-sinx
Hi noblerare! :smile:

Well, cos^2 + sin^2 = 1.

And sin2x = 2.sinx.cosx. :smile:

You really need to learn these formulas: :wink:

sin2x = 2.sinx.cosx = 2tanx.cos^2(x)
cos2x = cos^2(x) - sin^2(x)
1 + cos2x = 2cos^2(x)
1 - cos2x = 2sin^2(x)​
 

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