We're going to use our triangle substitution here. We can see it's an addition problem. So our variable is going to go on the opposite side so that our constants on the adjacent. So we get tangent Theta equaling X. So secant squared Theta D Theta equals DX. So then when we do this integration 8X, so eight tangent Theta over square root X ^2 square each side. So we can see that's tangent squared Theta plus one DX is going to be secant squared Theta D Theta. We're going to change our bounds. So if we put in tangent Theta equaling root 3, we can see that Theta equals tangent inverse of root 3 and that happens at π thirds and then tangent Theta equaling 0. So Theta equal tangent inverse of 0. That happens at 0. So now the square root of tangent squared Theta plus one is a secant Theta. So we get zero to π thirds. Eight could go on out in front. Tangent Theta, secant squared Theta, D Theta. The square root of secant squared Theta is going to be secant Theta. So this is going to turn into tangent Theta, secant Theta, D Theta, and that integrates to secant Theta. So then when we put in our bounds, we get 8 secant of π thirds -8 secant of 0. Secant of π thirds is 2, secant of 0 is 1. So we get 16 -, 8 or 8.