Integrating Arcsine: Solving for the Area Between Two Intervals

[tjbateh]

Homework Statement

[tex]\int[/tex][tex]^{1/\sqrt{2}}_{0}[/tex] [tex]\stackrel{arcsinx}{\sqrt{1-x^2}}[/tex]

Homework Equations

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

The Attempt at a Solution

My attempt was to set u= arcsinx, then DU would equal the bottom, so it would be [tex]\stackrel{u}{du}[/tex]...but from there I get stuck. I know the answer is supposed to be .308...Help anyone?

 

[Gregg]

[tex]\int_0^{1\over\sqrt{2}}\frac{\arcsin(x)}{\sqrt{1-x^2}}dx[/tex]

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

If you set u=arcsinx

[tex]du = \frac{dx}{\sqrt{1-x^2}}[/tex]

[tex]\int u du = \int {arcsinx\over \sqrt{1-x^2}} dx[/tex]

 

[tjbateh]

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

 

[Gregg]

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 won't need to change all back to x.

 

[tjbateh]

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

 

[Gregg]

[tex]\int_0^{1\over\sqrt{2}}\frac{\arcsin(x)}{\sqrt{1-x^2}}dx[/tex]

using the substition:

[tex] u = \arcsin(x) [/tex]

[tex] {du\over dx} = {1\over \sqrt{1-x^2}} \Rightarrow du = {dx\over \sqrt{1-x^2}} [/tex]

[tex] \int u du = \int \arcsinx {dx\over \sqrt{1-x^2}} [/tex]


then make sure the limits are changed:


[tex]x_0 = 0[/tex], [tex] x_1 = {1\over\sqrt{2}} [/tex]

[tex]\arcsin(0) = u_0 = 0[/tex]

[tex]\arcsin({1\over\sqrt{2}}) = u_1 = {\pi\over 4}[/tex]

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

 

[tjbateh]

ok so


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


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

 

[tjbateh]

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

 

[Gregg]

No. The whole integral has been transformed.

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

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

 

[tjbateh]

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

 

[Gregg]

integrate u from 0 to pi/4 and that's it.

 

[tjbateh]

so I would get .903??

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

 

[tjbateh]

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

[tex]\int^{\frac{1}{\sqrt{2}}}_{0}[/tex] [tex]\frac{1}{2}[/tex] arcsin²x


then from there they get [tex]\frac{\Pi^2}{32}[/tex]


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