Trig integrating with absolute values:

[Zeth]

Homework Statement

$$\int_0 ^\pi \sqrt{1-\sin^2 x} dx$$

Homework Equations

$$1 - \sin^2 x = \cos^2 x$$

The Attempt at a Solution

I don't know how to treat this since cos changes sign half way across the integral. I know the answer should be 2 but I keep getting 0 every which way I try.

 

[bel]

$$\int_0 ^\pi \sqrt{1-\sin^2 x} dx= \int_0 ^\pi |{\cos (x)}| dx$$

So you find where the $$0= \cos (x)$$ on the interval $$(0,\pi)$$ and then integrate separately.

i.e, $$\int_0 ^\pi |{\cos (x)}| dx = \int_0 ^{\frac{\pi}{2} }\cos (x) dx - \int_{\frac{\pi}{2}}^\pi \cos (x) dx$$

 

[neutrino]

You can split the integral over the two intervals, can't you?

 

[Gib Z]

I know the answer should be 2 but I keep getting 0 every which way I try.

Are you sure :)? Certainly the area is 2, but the integral gives you the Signed area. As neutrino said, splitting the integral into 2 gives the area :), as long as you take the absolute value of the negative value you will get from the integral between pi/2 and pi.