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**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?