Integration bu inverse trigonometric substitution Use an appropriate substitution to find the value of the following definite integral 2dr? 3 1! da: 0 V352 +4 (Note: in order to input inverse trigonometric functions you will need to type arcsin(x) , arccos (x), or arctan(:) as apprapriate). The appropriate Substitution is u = equivalentlg :1: 2 Using a reference triangle cos[u) = The differential is is: 2 tin. When integrating with respect to this n, the new limits of integration are (lower limit) and (upper limit}. The value of the integral is 2 as a=3 da:= I} v$2+4