Help Solving Integral of sqrt(2 - x^2) dx

[-JammyDodger-]

I was absent last friday when we went through it in class, so I'm completely lost when it comes to solving it. If anyone could explain how to do them I'd be extremely greatful. Thanks.

Homework Statement

It said use the substitution x = a sin theta.

Homework Equations

I'm not sure how to write this, but here it goes:

The integral of (Top limit = 1; Bottom limit = 0) sqrt(2 - x^2) dx

The Attempt at a Solution

I let x = sqrt(2) sin theta; so dx = sqrt(2) cos theta dtheta.

Then I subbed what I know for x and dx into the original equation -

The integral of: sqrt(2 - 2 sin^2 theta) sqrt(2) cos theta dtheta.

Now I took out 2 from the equation, and put it ouside of the integral -

So now i have -

(2)Integral of: Sqrt(1 - sin^2 theta) cos theta dtheta

And I know that cos^2A + sin^2A = 1; So I can put cos^2 Theta into my equation -

Now I have -

(2)Integral of: sqrt(cos^2 theta) cos theta dtheta.

Removing squareroot, and mulitplying the cos thetas gives me -

(2)Integral of: 1/2(1 + cos2theta) dtheta

then...

Integral of: (1 + cos2theta) dtheta.

then integrating...

[theta + sin2theta/2]


And that's where I'm stuck, I probably did that completely wrong anyway.




Thanks.

P.S. You don't even have to answer this particular question, even if you could work through a similar kind so I could see what to do it'd be perfect.

Thanks.

 

[dirk_mec1]

use a = sqrt(2) and 1-sin²(x) = cos²(x)

 

[-JammyDodger-]

use a = sqrt(2) and 1-sin²(x) = cos²(x)

Thanks, I think I see where to go now. Would it be possible for someone to put a final answer up? So that way I'll know what to work towards (theres no final answers in the book I am using).

Thanks.

 

[HallsofIvy]

What you did in your first post is correct. All you need to do is either change your theta variable back to x or (better) change the limits of integration. With x= sqrt(2)sin(theta), when x= 1, 1= sqrt(2)sin(theta) so sin(theta)= 1/sqrt(2)= sqrt(2)/2. What is theta? When x= 0, 0= sqrt(2)sin(theta) so sin(theta)= 0. What is theta? Use those as your limits of integration.

 

[-JammyDodger-]

What you did in your first post is correct. All you need to do is either change your theta variable back to x or (better) change the limits of integration. With x= sqrt(2)sin(theta), when x= 1, 1= sqrt(2)sin(theta) so sin(theta)= 1/sqrt(2)= sqrt(2)/2. What is theta? When x= 0, 0= sqrt(2)sin(theta) so sin(theta)= 0. What is theta? Use those as your limits of integration.

Thanks for your reply, I think I have it now. Would you mind telling me what the correct answer is if it isn't too much trouble? See there are no answers on the sheet we were given. For my final answer I got (pi + 2)/4. Is this correct?

Thanks again.

 

[HallsofIvy]

Yes, that is correct.

 

[-JammyDodger-]

Yes, that is correct.

Thank you.