Definite integral with inverse trigonometric solution? Let's say you have some function that when integrated gives an inverse trig function as a solution. If we're evaluating a definite integral, then there will be an infinite number of solutions. But when graphing the original function, there's only going to be one area under it. So why is this area equal to (I'm assuming), the inverse trig function evaluated at the first solution? How are these algebraic and geometric interpretations reconciled?

In order to make inverses of trigonometric functions 'functions', the primary values of ranges are defined for each of them. Thus, for real numbers, inverse trigonometric functions output at most one value. For knowing principal ranges of inverse trigonometric functions, refer to this.