How Effective Is the Tan(x/2) Substitution for Solving ∫√(1-sin(x))dx?

[noblerare]

I'm looking to solve this integral with the tan(x/2) substitution but so far, I don't know what to do.

Homework Statement

$$\int$$$$\sqrt{1-sinx}dx$$

Homework Equations

u=tan(x/2)

The Attempt at a Solution

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

sinx=$$\frac{2u}{1+u^2}$$
dx=$$\frac{2du}{1+u^2}$$

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

Now I don't know what to do.

Another method I tried is multiplying by conjugate:

$$\int$$$$\sqrt{1-sinx}$$$$\frac{\sqrt{1+sinx}}{\sqrt{1+sinx}}$$dx

$$\int$$$$\frac{\sqrt{1-sin^2(x)}}{\sqrt{1+sinx}}$$dx

$$\int$$$$\frac{cosxdx}{\sqrt{1+sinx}}$$dx

u=sinx
du=cosxdx

$$\int$$$$\frac{du}{\sqrt{1+u}}$$

$$\int$$(u+1)^(-$$\frac{1}{2}$$)du

2(u+1)^($$\frac{1}{2}$$

2$$\sqrt{sinx+1}$$+ C

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

Which way should I approach this problem?

 

[tiny-tim]

sinx=$$\frac{2u}{1+u^2}$$
dx=$$\frac{2du}{1+u^2}$$

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

Hi noblerare!

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.

 

[sutupidmath]

}}[/tex]dx

$$\int$$$$\frac{\sqrt{1-sin^2(x)}}{\sqrt{1+sinx}}$$dx

$$\int$$$$\frac{cosxdx}{\sqrt{1+sinx}}$$dx



2$$\sqrt{sinx+1}$$+ 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?

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

If the answer in the book is different from this one, they should be equivalent, because nothing is wrong with this.

 

[noblerare]

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

 

[sutupidmath]

Noblerare,

Your answer is correct, what else are you looking for??

 

[tiny-tim]

… 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).

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!

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

And sin2x = 2.sinx.cosx.

You really need to learn these formulas:

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)​