# Integral of the function?

1. Oct 12, 2011

### wlooi

1. The problem statement, all variables and given/known data

∫(sin x)√(1+cos2x) dx

Last edited: Oct 12, 2011
2. Oct 12, 2011

### HallsofIvy

What you have written makes no sense (even if the latex were working). you have \sqrt{} with nothing inside the square root and have a subscript that makes no sense.

Did you mean $\int sin(x)\sqrt{1+ cos^2(x)}dx$?

If so, start by letting y= cos(x).

3. Oct 12, 2011

### wlooi

Yes, that's exactly what I was try to make. Well, this is my first time using this so sorry..

4. Oct 12, 2011

### wlooi

Well, I got myself until:

$\large -$ $\int \sqrt{1+y^2}$ $\large dy$

then I stucked again.

5. Oct 12, 2011

### Dick

Try and think of a trig (or hyperbolic trig) substitution that will turn 1+y^2 into the square of something so you can get rid of the square root.

6. Oct 12, 2011

### wlooi

Argh... I don't seem to get it after cracking my head for a while. Can I have a little bit more tips?

7. Oct 12, 2011

### SammyS

Staff Emeritus
Let's see, ...

1+tan2(θ) = sec2(θ)

1+sinh2(u) = cosh2(u)

8. Oct 14, 2011

### wlooi

using 1+tan2v=sec2v , I got it until :

-∫sec3v dv

am I suppose to do by parts or is there other ways?

9. Oct 14, 2011

### Dick

Parts. Split it into sec(v)^2*dv and sec(v).

10. Oct 14, 2011

### wlooi

Okay, I think I got it. Thanks for the help from everyone.^^