# Perplexing integration - sqrt(1+sinx)

1. Nov 16, 2012

### peripatein

Hello,

I am trying to integrate sqrt(1+sinx) dx but I keep getting 2sqrt(1-sinx) (having initially substituted u=sinx), which is wrong!! Could anyone please tell me what I am doing wrong?

2. Nov 16, 2012

### micromass

Staff Emeritus
We can't really know what you did wrong unless you tell us what you did!! So, can you show us your work?

3. Nov 16, 2012

### peripatein

As I wrote, I substituted u=sinx and got that the integral should be equal to 2sqrt(1-sinx). Which is wrong, according to Wolfram and a simple differentiation test. I am simply not sure where I faltered. May you please advise?

4. Nov 16, 2012

### dextercioby

You're starting in the wrong place. You should be using a different substitution. Try tan x/2 = u.

5. Nov 16, 2012

### peripatein

I would have thought so, except that I was instructed to use that substitution. :s
It should still be correct, granted the algebra is. Yet the answer is incorrect!

6. Nov 16, 2012

### D H

Staff Emeritus

7. Nov 16, 2012

### peripatein

u=sinx, du=cosxdx
Hence, int=[sqrt(1+u)/sqrt(1-u^2)] du=[1/sqrt(1-u)] du.
Hence, int=2sqrt(1-sinx)

8. Nov 16, 2012

### D H

Staff Emeritus
Try that last step again.

9. Nov 16, 2012

### peripatein

What do you mean? Dividing by sqrt(1+u)?

10. Nov 16, 2012

### SammyS

Staff Emeritus
Do the integration:

$\displaystyle \int \frac{1}{\sqrt{1-u}}\,du\ .$

11. Nov 16, 2012

### peripatein

But isn't that 2sqrt(1-u)?

12. Nov 16, 2012

### SammyS

Staff Emeritus
sign error.

13. Nov 16, 2012

### peripatein

14. Nov 16, 2012

### peripatein

Wrong link. In any case, that is NOT the answer. Question is, why?

15. Nov 16, 2012

### D H

Staff Emeritus
They're the same answer *if* $\cos x=\sqrt{1-\sin^2 x}$. That's not always the case, but you implicitly assumed that this is an identity in your u-substitution.

16. Nov 16, 2012

### SammyS

Staff Emeritus

You have to be careful about sign behavior of the functions you're working with.

$\displaystyle \cos(x)=\sqrt{1-\sin^2(x)}\ \$ only when cos(x) ≥ 0.

Otherwise, $\displaystyle \ \cos(x)=-\sqrt{1-\sin^2(x)\ .}$

17. Nov 16, 2012

### peripatein

Okay, thanks!
I am also trying to evaluate the improper integral 1/[(x-1)^1/3] between 0 and 4. May I do so thus?:

{ lim b->1- int 1/[(x-1)^1/3] dx between 0 and b } + { lim b->1+ int 1/[(x-1)^1/3] dx between b and 4 }?

Is there a another, better way to evaluate it?

18. Nov 16, 2012

### SammyS

Staff Emeritus
It's best to start a new thread for this.

It's a completely different problem.