1.7 Integrals Resulting in Inverse Trigonometric Functions [STARTS AT 16:13] PROFESSOR: Evaluate from 0 to 2 dx over 4 plus x squared. So a equals 2 squared there. u equals x. So du equals dx. Our limits of integration are going to stay the same because u is x. We're not changing them there. Over a squared plus u squared. This is arctan. Now, because a is 2, we're going have a factor of 1/2. And we're also going to say u divided by a, which is 2. And we are going to evaluate this from 0 to 2. So plugging a 2 in there, that's going to be 0.5 arctan of 1, which is pi over 8. Plugging 0 in there-- 0.5 arctan of 0. So pi over 8 is our answer there. [ENDS AT 17:34]