When doing this kind of problem, we're going to use triangle substitution. We can see that it's a plus here. So we're going to have our variable be on the opposite side and our constant be on the adjacent so that we have tangent Theta equaling X / 10. So 10 tangent Theta is equaling X 10 secant squared Theta D Theta equal DX. So when we substitute the DX is going to get replaced by 10 secant squared Theta D Theta. The denominator is going to turn into 100 + X ^2 while the X was 10 secant Theta. So X ^2 if we square each side would be 100 tangent squared Theta. Now, changing our bounds, if we stick 10 root 3IN for our X, we can see that we want tangent Theta equaling root 3, so theta's tangent inverse of root 3. And we know that that occurs at π thirds. Doing the same thing for the bottom, we get tangent Theta equaling. We get Theta equaling tangent inverse of negative root 3 if we do the bottom bounds. And that's going to occur at negative π thirds. So now when we look at this, the 10 / 100, I factor out the 10 over the 100. That leaves US1 plus tangent squared Theta, which we know by our Pythagorean identities is just secant squared Theta. So the secant squared Theta on top and bottom reduced the 10 / 100 reduced to 110th. So this really turns into 110th Theta from π thirds to negative π thirds. So when we have 110th Pi third minus negative Pi third is going to be 1/10 * 2π thirds or just π fifteenths.