# Sub in the two intervals?

1. Sep 21, 2009

### tjbateh

1. The problem statement, all variables and given/known data
$$\int$$$$^{1/\sqrt{2}}_{0}$$ $$\stackrel{arcsinx}{\sqrt{1-x^2}}$$

2. Relevant equations

The 0 is supposed to be on the bottom of the intergal, but I could not format it to go there.

3. The attempt at a solution
My attempt was to set u= arcsinx, then DU would equal the bottom, so it would be $$\stackrel{u}{du}$$...but from there I get stuck. I know the answer is supposed to be .308...Help anyone?

Last edited by a moderator: Jun 1, 2014
2. Sep 21, 2009

### Gregg

Re: Integration

$$\int_0^{1\over\sqrt{2}}\frac{\arcsin(x)}{\sqrt{1-x^2}}dx$$

\int_0^{1\over\sqrt{2}}\frac{\arcsin(x)}{\sqrt{1-x^2}}

If you set u=arcsinx

$$du = \frac{dx}{\sqrt{1-x^2}}$$

$$\int u du = \int {arcsinx\over \sqrt{1-x^2}} dx$$

3. Sep 21, 2009

### tjbateh

Re: Integration

Ok, that makes sense. Then from there do you just sub in the two intervals?

4. Sep 21, 2009

### Gregg

Re: Integration

Yeah you need to make sure the intervals are consistent i.e. u=arcsinx, sinu=0 or 1/sqrt(2) quite simple, and that way you wont need to change all back to x.

5. Sep 21, 2009

### tjbateh

Re: Integration

If possible, could you set up the substitution for me so I can see how it looks, and then I can work from that?

6. Sep 21, 2009

### Gregg

Re: Integration

$$\int_0^{1\over\sqrt{2}}\frac{\arcsin(x)}{\sqrt{1-x^2}}dx$$

using the substition:

$$u = \arcsin(x)$$

$${du\over dx} = {1\over \sqrt{1-x^2}} \Rightarrow du = {dx\over \sqrt{1-x^2}}$$

$$\int u du = \int \arcsinx {dx\over \sqrt{1-x^2}}$$

then make sure the limits are changed:

$$x_0 = 0$$, $$x_1 = {1\over\sqrt{2}}$$

$$\arcsin(0) = u_0 = 0$$

$$\arcsin({1\over\sqrt{2}}) = u_1 = {\pi\over 4}$$

$$\int_0^{1\over \sqrt{2}} {\arcsin(x)\over \sqrt{1-x^2}} dx = \int_0^{\pi\over 4} u du$$

7. Sep 21, 2009

### tjbateh

Re: Integration

ok so

$$\int^{\Pi/4}_{0}$$ $$dx/\sqrt{1-x^2}$$

Then from there I substitute pie/4 in for x?

8. Sep 21, 2009

### tjbateh

Re: Integration

or would it be something like this..arcsin (Pie/4) - arcsin (0)??

9. Sep 21, 2009

### Gregg

Re: Integration

No. The whole integral has been transformed.

$$\int_{0}^{\pi\over 4} u du = \int_{0}^{1\over\sqrt{2}} \frac{\arcsin (x)}{\sqrt{1-x^2}} dx$$

Basically because you swapped x for sin(u) (essentially) if x is $$1\over\sqrt{2}$$ then u has to be $$\pi\over 4$$ doesn't it? So everything is consistent? 0 stays the same because sin(0)=0.

10. Sep 21, 2009

### tjbateh

Re: Integration

True, so we have that set up. Now where do we go from there?

11. Sep 21, 2009

### Gregg

Re: Integration

integrate u from 0 to pi/4 and thats it.

12. Sep 21, 2009

### tjbateh

Re: Integration

so I would get .903??

The answer in the back of the book for this problem says .308.
Did I do something wrong?

13. Sep 22, 2009

### tjbateh

Re: Integration

on calcchat.com, they have something like this for their new integral.

$$\int^{\frac{1}{\sqrt{2}}}_{0}$$ $$\frac{1}{2}$$ arcsin2x

then from there they get $$\frac{\Pi^2}{32}$$

which approximates to 0.308, but how on earth did they get that new integral? Does what you see here make any sense?