- #1
noblerare
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I'm looking to solve this integral with the tan(x/2) substitution but so far, I don't know what to do.
[tex]\int[/tex][tex]\sqrt{1-sinx}dx[/tex]
u=tan(x/2)
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?
Homework Statement
[tex]\int[/tex][tex]\sqrt{1-sinx}dx[/tex]
Homework Equations
u=tan(x/2)
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?