# Converting degree into 'x' (integration)

1. Nov 26, 2007

### L²Cc

1. The problem statement, all variables and given/known data
After integrating x²/(sqrt1-x²) using the triangle method and thereby substituting sin¤ for x, I ended up with the result 1/2¤-1/2sin2¤+c. For my answer toge complete, I must have the ¤ in x format. I managed to convert the 1/2¤, which is simply 1/2arcsinx but I couldn't for 1/2sin2¤, my answer did not agree with the textbook's which read 1/2x(sqrt1-x²)....

My computer won't read latex at the moment so please excuse the amateur notation. Thank you.

2. Relevant equations
I think I covered it all above!

2. Nov 26, 2007

### Dick

If I call your substitution sin(t) (I don't know what symbol you were trying to put in), sin(2t)=2*cos(t)*sin(t). Put t=arcsin(x) in there.

3. Nov 26, 2007

### L²Cc

Oh I see, you used the trig identity....and then I refer back to the triangle to figure out what's cos(t)...Thank you!

4. Nov 26, 2007

### L²Cc

Following your reasoning, 1/2*sin(2t) becomes 2x*sqrt(1-x²). The two cancels out when I plug it back into 1/2*sin(2t), and this can't be since the answer is 1/2*x*sqrt(1-x²)??!

5. Nov 26, 2007

### Dick

I get that the integral of sin(t)^2 is t/2-sin(2t)/4. You are missing a factor of 2.

6. Nov 26, 2007

### L²Cc

So you're suggesting that the integral of 1/2cos(2t) is 1/2sin(2t) ?! (can you please elaborate as to how you came up with the answer?)

7. Nov 26, 2007

### Dick

No, I'm suggesting it's sin(2t)/4. Substitute u=2t, du=2dt. That's your missing 2.

8. Nov 26, 2007

### L²Cc

hmm I see, I never did that before, I simply went on with the integrating whenever it was 1/2-1/2cos2t for example....and I always obtained the right answer, weird! But thank you!