00:01
So i began this problem by separating my variables, and all i did was multiply both sides of my equation by dx, and then divide by tangent of y.
00:09
And so from here, we're going to be able to integrate both sides of our equation.
00:14
So i'm actually going to rewrite this left side as the integral of 1 over, or excuse me, not 1 over, but cosine y over, d.
00:23
And that's going to allow me to use u substitution here.
00:26
So i'll take you to be equal to sine y, which then i get du is equal to cosine of y, d .y.
00:35
And so my left hand side of my equation is going to become the integral of 1 over u, du, and it'll be equal to the integral of negative 2x dx...
