Perplexing integration - sqrt(1+sinx)

[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?

 

[micromass]

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

 

[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?

 

[dextercioby]

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

 

[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!

 

[D H]

peripatein, you need to show us your work. We cannot read your mind.

 

[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)

 

[D H]

Try that last step again.

 

[peripatein]

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

 

[SammyS]

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

Do the integration:

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

 

[peripatein]

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

 

[SammyS]

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

sign error.

 

[peripatein]

Correct, it should be -2sqrt(1-u), but that is still miles away from the true answer, which is
http://www.wolframalpha.com/widget/...ds: Indefinite Integral Calculator&theme=blue

 

[peripatein]

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

 

[D H]

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.

 

[SammyS]

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

What is the correct answer?

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

 

[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?

 

[SammyS]

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?

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

It's a completely different problem.